Javascript: array plus number [duplicate] - javascript

This question already has answers here:
What is the explanation for these bizarre JavaScript behaviours mentioned in the 'Wat' talk for CodeMash 2012?
(5 answers)
Closed 9 years ago.
Some operations in javascript returns unexpected results. One is extremely strange:
[] + 1 = "1"
Can anybody explain why it works like that?

[] are converted to an empty string due to + operator. so "" + 1 => "1" (number converted to string too)

Javascript's rules for addition between differeent types are as follows:
Given the following addition.
value1 + value2
To evaluate this expression, the following steps are taken (§11.6.1):
Convert both operands to primitives (mathematical notation, not JavaScript):
prim1 := ToPrimitive(value1)
prim2 := ToPrimitive(value2)
PreferredType is omitted and thus Number for non-dates, String for dates.
If either prim1 or prim2 is a string then convert both to strings and return the concatenation of the results.
Otherwise, convert both prim1 and prim2 to numbers and return the sum of the results.
Source
In this case the array gets converted to an empty string, and then the + performs string concatenation

ECMAScript 11.6.1 defines addition. Steps 5 and 6 of addition call ToPrimitive (9.1) for each operand and operate on those results:
For an array (or any object), ToPrimative calls the toString method of the object. The result of calling toString on an empty array is the empty string (per the behavior described in 15.4.4.2.
For a number, ToPrimitive returns the number (since numbers are already primitive).
We're left adding the empty string and the number 1. When either operand in addition is a string, the addition acts as a concatenation operation (per addition's step 7), so we end up with "" + "1" = "1".

Related

why is []+[] equal to an empty string in javascript [duplicate]

The 'Wat' talk for CodeMash 2012 basically points out a few bizarre quirks with Ruby and JavaScript.
I have made a JSFiddle of the results at http://jsfiddle.net/fe479/9/.
The behaviours specific to JavaScript (as I don't know Ruby) are listed below.
I found in the JSFiddle that some of my results didn't correspond with those in the video, and I am not sure why. I am, however, curious to know how JavaScript is handling working behind the scenes in each case.
Empty Array + Empty Array
[] + []
result:
<Empty String>
I am quite curious about the + operator when used with arrays in JavaScript.
This matches the video's result.
Empty Array + Object
[] + {}
result:
[Object]
This matches the video's result. What's going on here? Why is this an object. What does the + operator do?
Object + Empty Array
{} + []
result:
[Object]
This doesn't match the video. The video suggests that the result is 0, whereas I get [Object].
Object + Object
{} + {}
result:
[Object][Object]
This doesn't match the video either, and how does outputting a variable result in two objects? Maybe my JSFiddle is wrong.
Array(16).join("wat" - 1)
result:
NaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
Doing wat + 1 results in wat1wat1wat1wat1...
I suspect this is just straightforward behaviour that trying to subtract a number from a string results in NaN.
Here's a list of explanations for the results you're seeing (and supposed to be seeing). The references I'm using are from the ECMA-262 standard.
[] + []
When using the addition operator, both the left and right operands are converted to primitives first (§11.6.1). As per §9.1, converting an object (in this case an array) to a primitive returns its default value, which for objects with a valid toString() method is the result of calling object.toString() (§8.12.8). For arrays this is the same as calling array.join() (§15.4.4.2). Joining an empty array results in an empty string, so step #7 of the addition operator returns the concatenation of two empty strings, which is the empty string.
[] + {}
Similar to [] + [], both operands are converted to primitives first. For "Object objects" (§15.2), this is again the result of calling object.toString(), which for non-null, non-undefined objects is "[object Object]" (§15.2.4.2).
{} + []
The {} here is not parsed as an object, but instead as an empty block (§12.1, at least as long as you're not forcing that statement to be an expression, but more about that later). The return value of empty blocks is empty, so the result of that statement is the same as +[]. The unary + operator (§11.4.6) returns ToNumber(ToPrimitive(operand)). As we already know, ToPrimitive([]) is the empty string, and according to §9.3.1, ToNumber("") is 0.
{} + {}
Similar to the previous case, the first {} is parsed as a block with empty return value. Again, +{} is the same as ToNumber(ToPrimitive({})), and ToPrimitive({}) is "[object Object]" (see [] + {}). So to get the result of +{}, we have to apply ToNumber on the string "[object Object]". When following the steps from §9.3.1, we get NaN as a result:
If the grammar cannot interpret the String as an expansion of StringNumericLiteral, then the result of ToNumber is NaN.
Array(16).join("wat" - 1)
As per §15.4.1.1 and §15.4.2.2, Array(16) creates a new array with length 16. To get the value of the argument to join, §11.6.2 steps #5 and #6 show that we have to convert both operands to a number using ToNumber. ToNumber(1) is simply 1 (§9.3), whereas ToNumber("wat") again is NaN as per §9.3.1. Following step 7 of §11.6.2, §11.6.3 dictates that
If either operand is NaN, the result is NaN.
So the argument to Array(16).join is NaN. Following §15.4.4.5 (Array.prototype.join), we have to call ToString on the argument, which is "NaN" (§9.8.1):
If m is NaN, return the String "NaN".
Following step 10 of §15.4.4.5, we get 15 repetitions of the concatenation of "NaN" and the empty string, which equals the result you're seeing.
When using "wat" + 1 instead of "wat" - 1 as argument, the addition operator converts 1 to a string instead of converting "wat" to a number, so it effectively calls Array(16).join("wat1").
As to why you're seeing different results for the {} + [] case: When using it as a function argument, you're forcing the statement to be an ExpressionStatement, which makes it impossible to parse {} as empty block, so it's instead parsed as an empty object literal.
This is more of a comment than an answer, but for some reason I can't comment on your question. I wanted to correct your JSFiddle code. However, I posted this on Hacker News and someone suggested that I repost it here.
The problem in the JSFiddle code is that ({}) (opening braces inside of parentheses) is not the same as {} (opening braces as the start of a line of code). So when you type out({} + []) you are forcing the {} to be something which it is not when you type {} + []. This is part of the overall 'wat'-ness of Javascript.
The basic idea was simple JavaScript wanted to allow both of these forms:
if (u)
v;
if (x) {
y;
z;
}
To do so, two interpretations were made of the opening brace: 1. it is not required and 2. it can appear anywhere.
This was a wrong move. Real code doesn't have an opening brace appearing in the middle of nowhere, and real code also tends to be more fragile when it uses the first form rather than the second. (About once every other month at my last job, I'd get called to a coworker's desk when their modifications to my code weren't working, and the problem was that they'd added a line to the "if" without adding curly braces. I eventually just adopted the habit that the curly braces are always required, even when you're only writing one line.)
Fortunately in many cases eval() will replicate the full wat-ness of JavaScript. The JSFiddle code should read:
function out(code) {
function format(x) {
return typeof x === "string" ?
JSON.stringify(x) : x;
}
document.writeln('>>> ' + code);
document.writeln(format(eval(code)));
}
document.writeln("<pre>");
out('[] + []');
out('[] + {}');
out('{} + []');
out('{} + {}');
out('Array(16).join("wat" + 1)');
out('Array(16).join("wat - 1")');
out('Array(16).join("wat" - 1) + " Batman!"');
document.writeln("</pre>");
[Also that is the first time I have written document.writeln in many many many years, and I feel a little dirty writing anything involving both document.writeln() and eval().]
I second #Ventero’s solution. If you want to, you can go into more detail as to how + converts its operands.
First step (§9.1): convert both operands to primitives (primitive values are undefined, null, booleans, numbers, strings; all other values are objects, including arrays and functions). If an operand is already primitive, you are done. If not, it is an object obj and the following steps are performed:
Call obj.valueOf(). If it returns a primitive, you are done. Direct instances of Object and arrays return themselves, so you are not done yet.
Call obj.toString(). If it returns a primitive, you are done. {} and [] both return a string, so you are done.
Otherwise, throw a TypeError.
For dates, step 1 and 2 are swapped. You can observe the conversion behavior as follows:
var obj = {
valueOf: function () {
console.log("valueOf");
return {}; // not a primitive
},
toString: function () {
console.log("toString");
return {}; // not a primitive
}
}
Interaction (Number() first converts to primitive then to number):
> Number(obj)
valueOf
toString
TypeError: Cannot convert object to primitive value
Second step (§11.6.1): If one of the operands is a string, the other operand is also converted to string and the result is produced by concatenating two strings. Otherwise, both operands are converted to numbers and the result is produced by adding them.
More detailed explanation of the conversion process: “What is {} + {} in JavaScript?”
We may refer to the specification and that's great and most accurate, but most of the cases can also be explained in a more comprehensible way with the following statements:
+ and - operators work only with primitive values. More specifically +(addition) works with either strings or numbers, and +(unary) and -(subtraction and unary) works only with numbers.
All native functions or operators that expect primitive value as argument, will first convert that argument to desired primitive type. It is done with valueOf or toString, which are available on any object. That's the reason why such functions or operators don't throw errors when invoked on objects.
So we may say that:
[] + [] is same as String([]) + String([]) which is same as '' + ''. I mentioned above that +(addition) is also valid for numbers, but there is no valid number representation of an array in JavaScript, so addition of strings is used instead.
[] + {} is same as String([]) + String({}) which is same as '' + '[object Object]'
{} + []. This one deserves more explanation (see Ventero answer). In that case, curly braces are treated not as an object but as an empty block, so it turns out to be same as +[]. Unary + works only with numbers, so the implementation tries to get a number out of []. First it tries valueOf which in the case of arrays returns the same object, so then it tries the last resort: conversion of a toString result to a number. We may write it as +Number(String([])) which is same as +Number('') which is same as +0.
Array(16).join("wat" - 1) subtraction - works only with numbers, so it's the same as: Array(16).join(Number("wat") - 1), as "wat" can't be converted to a valid number. We receive NaN, and any arithmetic operation on NaN results with NaN, so we have: Array(16).join(NaN).
To buttress what has been shared earlier.
The underlying cause of this behaviour is partly due to the weakly-typed nature of JavaScript. For example, the expression 1 + “2” is ambiguous since there are two possible interpretations based on the operand types (int, string) and (int int):
User intends to concatenate two strings, result: “12”
User intends to add two numbers, result: 3
Thus with varying input types,the output possibilities increase.
The addition algorithm
Coerce operands to primitive values
The JavaScript primitives are string, number, null, undefined and boolean (Symbol is coming soon in ES6). Any other value is an object (e.g. arrays, functions and objects). The coercion process for converting objects into primitive values is described thus:
If a primitive value is returned when object.valueOf() is invoked, then return this value, otherwise continue
If a primitive value is returned when object.toString() is invoked, then return this value, otherwise continue
Throw a TypeError
Note: For date values, the order is to invoke toString before valueOf.
If any operand value is a string, then do a string concatenation
Otherwise, convert both operands to their numeric value and then add these values
Knowing the various coercion values of types in JavaScript does help to make the confusing outputs clearer. See the coercion table below
+-----------------+-------------------+---------------+
| Primitive Value | String value | Numeric value |
+-----------------+-------------------+---------------+
| null | “null” | 0 |
| undefined | “undefined” | NaN |
| true | “true” | 1 |
| false | “false” | 0 |
| 123 | “123” | 123 |
| [] | “” | 0 |
| {} | “[object Object]” | NaN |
+-----------------+-------------------+---------------+
It is also good to know that JavaScript's + operator is left-associative as this determines what the output will be cases involving more than one + operation.
Leveraging the
Thus 1 + "2" will give "12" because any addition involving a string will always default to string concatenation.
You can read more examples in this blog post (disclaimer I wrote it).

Converting Between Types JavaScript [duplicate]

The 'Wat' talk for CodeMash 2012 basically points out a few bizarre quirks with Ruby and JavaScript.
I have made a JSFiddle of the results at http://jsfiddle.net/fe479/9/.
The behaviours specific to JavaScript (as I don't know Ruby) are listed below.
I found in the JSFiddle that some of my results didn't correspond with those in the video, and I am not sure why. I am, however, curious to know how JavaScript is handling working behind the scenes in each case.
Empty Array + Empty Array
[] + []
result:
<Empty String>
I am quite curious about the + operator when used with arrays in JavaScript.
This matches the video's result.
Empty Array + Object
[] + {}
result:
[Object]
This matches the video's result. What's going on here? Why is this an object. What does the + operator do?
Object + Empty Array
{} + []
result:
[Object]
This doesn't match the video. The video suggests that the result is 0, whereas I get [Object].
Object + Object
{} + {}
result:
[Object][Object]
This doesn't match the video either, and how does outputting a variable result in two objects? Maybe my JSFiddle is wrong.
Array(16).join("wat" - 1)
result:
NaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
Doing wat + 1 results in wat1wat1wat1wat1...
I suspect this is just straightforward behaviour that trying to subtract a number from a string results in NaN.
Here's a list of explanations for the results you're seeing (and supposed to be seeing). The references I'm using are from the ECMA-262 standard.
[] + []
When using the addition operator, both the left and right operands are converted to primitives first (§11.6.1). As per §9.1, converting an object (in this case an array) to a primitive returns its default value, which for objects with a valid toString() method is the result of calling object.toString() (§8.12.8). For arrays this is the same as calling array.join() (§15.4.4.2). Joining an empty array results in an empty string, so step #7 of the addition operator returns the concatenation of two empty strings, which is the empty string.
[] + {}
Similar to [] + [], both operands are converted to primitives first. For "Object objects" (§15.2), this is again the result of calling object.toString(), which for non-null, non-undefined objects is "[object Object]" (§15.2.4.2).
{} + []
The {} here is not parsed as an object, but instead as an empty block (§12.1, at least as long as you're not forcing that statement to be an expression, but more about that later). The return value of empty blocks is empty, so the result of that statement is the same as +[]. The unary + operator (§11.4.6) returns ToNumber(ToPrimitive(operand)). As we already know, ToPrimitive([]) is the empty string, and according to §9.3.1, ToNumber("") is 0.
{} + {}
Similar to the previous case, the first {} is parsed as a block with empty return value. Again, +{} is the same as ToNumber(ToPrimitive({})), and ToPrimitive({}) is "[object Object]" (see [] + {}). So to get the result of +{}, we have to apply ToNumber on the string "[object Object]". When following the steps from §9.3.1, we get NaN as a result:
If the grammar cannot interpret the String as an expansion of StringNumericLiteral, then the result of ToNumber is NaN.
Array(16).join("wat" - 1)
As per §15.4.1.1 and §15.4.2.2, Array(16) creates a new array with length 16. To get the value of the argument to join, §11.6.2 steps #5 and #6 show that we have to convert both operands to a number using ToNumber. ToNumber(1) is simply 1 (§9.3), whereas ToNumber("wat") again is NaN as per §9.3.1. Following step 7 of §11.6.2, §11.6.3 dictates that
If either operand is NaN, the result is NaN.
So the argument to Array(16).join is NaN. Following §15.4.4.5 (Array.prototype.join), we have to call ToString on the argument, which is "NaN" (§9.8.1):
If m is NaN, return the String "NaN".
Following step 10 of §15.4.4.5, we get 15 repetitions of the concatenation of "NaN" and the empty string, which equals the result you're seeing.
When using "wat" + 1 instead of "wat" - 1 as argument, the addition operator converts 1 to a string instead of converting "wat" to a number, so it effectively calls Array(16).join("wat1").
As to why you're seeing different results for the {} + [] case: When using it as a function argument, you're forcing the statement to be an ExpressionStatement, which makes it impossible to parse {} as empty block, so it's instead parsed as an empty object literal.
This is more of a comment than an answer, but for some reason I can't comment on your question. I wanted to correct your JSFiddle code. However, I posted this on Hacker News and someone suggested that I repost it here.
The problem in the JSFiddle code is that ({}) (opening braces inside of parentheses) is not the same as {} (opening braces as the start of a line of code). So when you type out({} + []) you are forcing the {} to be something which it is not when you type {} + []. This is part of the overall 'wat'-ness of Javascript.
The basic idea was simple JavaScript wanted to allow both of these forms:
if (u)
v;
if (x) {
y;
z;
}
To do so, two interpretations were made of the opening brace: 1. it is not required and 2. it can appear anywhere.
This was a wrong move. Real code doesn't have an opening brace appearing in the middle of nowhere, and real code also tends to be more fragile when it uses the first form rather than the second. (About once every other month at my last job, I'd get called to a coworker's desk when their modifications to my code weren't working, and the problem was that they'd added a line to the "if" without adding curly braces. I eventually just adopted the habit that the curly braces are always required, even when you're only writing one line.)
Fortunately in many cases eval() will replicate the full wat-ness of JavaScript. The JSFiddle code should read:
function out(code) {
function format(x) {
return typeof x === "string" ?
JSON.stringify(x) : x;
}
document.writeln('>>> ' + code);
document.writeln(format(eval(code)));
}
document.writeln("<pre>");
out('[] + []');
out('[] + {}');
out('{} + []');
out('{} + {}');
out('Array(16).join("wat" + 1)');
out('Array(16).join("wat - 1")');
out('Array(16).join("wat" - 1) + " Batman!"');
document.writeln("</pre>");
[Also that is the first time I have written document.writeln in many many many years, and I feel a little dirty writing anything involving both document.writeln() and eval().]
I second #Ventero’s solution. If you want to, you can go into more detail as to how + converts its operands.
First step (§9.1): convert both operands to primitives (primitive values are undefined, null, booleans, numbers, strings; all other values are objects, including arrays and functions). If an operand is already primitive, you are done. If not, it is an object obj and the following steps are performed:
Call obj.valueOf(). If it returns a primitive, you are done. Direct instances of Object and arrays return themselves, so you are not done yet.
Call obj.toString(). If it returns a primitive, you are done. {} and [] both return a string, so you are done.
Otherwise, throw a TypeError.
For dates, step 1 and 2 are swapped. You can observe the conversion behavior as follows:
var obj = {
valueOf: function () {
console.log("valueOf");
return {}; // not a primitive
},
toString: function () {
console.log("toString");
return {}; // not a primitive
}
}
Interaction (Number() first converts to primitive then to number):
> Number(obj)
valueOf
toString
TypeError: Cannot convert object to primitive value
Second step (§11.6.1): If one of the operands is a string, the other operand is also converted to string and the result is produced by concatenating two strings. Otherwise, both operands are converted to numbers and the result is produced by adding them.
More detailed explanation of the conversion process: “What is {} + {} in JavaScript?”
We may refer to the specification and that's great and most accurate, but most of the cases can also be explained in a more comprehensible way with the following statements:
+ and - operators work only with primitive values. More specifically +(addition) works with either strings or numbers, and +(unary) and -(subtraction and unary) works only with numbers.
All native functions or operators that expect primitive value as argument, will first convert that argument to desired primitive type. It is done with valueOf or toString, which are available on any object. That's the reason why such functions or operators don't throw errors when invoked on objects.
So we may say that:
[] + [] is same as String([]) + String([]) which is same as '' + ''. I mentioned above that +(addition) is also valid for numbers, but there is no valid number representation of an array in JavaScript, so addition of strings is used instead.
[] + {} is same as String([]) + String({}) which is same as '' + '[object Object]'
{} + []. This one deserves more explanation (see Ventero answer). In that case, curly braces are treated not as an object but as an empty block, so it turns out to be same as +[]. Unary + works only with numbers, so the implementation tries to get a number out of []. First it tries valueOf which in the case of arrays returns the same object, so then it tries the last resort: conversion of a toString result to a number. We may write it as +Number(String([])) which is same as +Number('') which is same as +0.
Array(16).join("wat" - 1) subtraction - works only with numbers, so it's the same as: Array(16).join(Number("wat") - 1), as "wat" can't be converted to a valid number. We receive NaN, and any arithmetic operation on NaN results with NaN, so we have: Array(16).join(NaN).
To buttress what has been shared earlier.
The underlying cause of this behaviour is partly due to the weakly-typed nature of JavaScript. For example, the expression 1 + “2” is ambiguous since there are two possible interpretations based on the operand types (int, string) and (int int):
User intends to concatenate two strings, result: “12”
User intends to add two numbers, result: 3
Thus with varying input types,the output possibilities increase.
The addition algorithm
Coerce operands to primitive values
The JavaScript primitives are string, number, null, undefined and boolean (Symbol is coming soon in ES6). Any other value is an object (e.g. arrays, functions and objects). The coercion process for converting objects into primitive values is described thus:
If a primitive value is returned when object.valueOf() is invoked, then return this value, otherwise continue
If a primitive value is returned when object.toString() is invoked, then return this value, otherwise continue
Throw a TypeError
Note: For date values, the order is to invoke toString before valueOf.
If any operand value is a string, then do a string concatenation
Otherwise, convert both operands to their numeric value and then add these values
Knowing the various coercion values of types in JavaScript does help to make the confusing outputs clearer. See the coercion table below
+-----------------+-------------------+---------------+
| Primitive Value | String value | Numeric value |
+-----------------+-------------------+---------------+
| null | “null” | 0 |
| undefined | “undefined” | NaN |
| true | “true” | 1 |
| false | “false” | 0 |
| 123 | “123” | 123 |
| [] | “” | 0 |
| {} | “[object Object]” | NaN |
+-----------------+-------------------+---------------+
It is also good to know that JavaScript's + operator is left-associative as this determines what the output will be cases involving more than one + operation.
Leveraging the
Thus 1 + "2" will give "12" because any addition involving a string will always default to string concatenation.
You can read more examples in this blog post (disclaimer I wrote it).

What is the result of sum of empty object and empty array in javascript? [duplicate]

The 'Wat' talk for CodeMash 2012 basically points out a few bizarre quirks with Ruby and JavaScript.
I have made a JSFiddle of the results at http://jsfiddle.net/fe479/9/.
The behaviours specific to JavaScript (as I don't know Ruby) are listed below.
I found in the JSFiddle that some of my results didn't correspond with those in the video, and I am not sure why. I am, however, curious to know how JavaScript is handling working behind the scenes in each case.
Empty Array + Empty Array
[] + []
result:
<Empty String>
I am quite curious about the + operator when used with arrays in JavaScript.
This matches the video's result.
Empty Array + Object
[] + {}
result:
[Object]
This matches the video's result. What's going on here? Why is this an object. What does the + operator do?
Object + Empty Array
{} + []
result:
[Object]
This doesn't match the video. The video suggests that the result is 0, whereas I get [Object].
Object + Object
{} + {}
result:
[Object][Object]
This doesn't match the video either, and how does outputting a variable result in two objects? Maybe my JSFiddle is wrong.
Array(16).join("wat" - 1)
result:
NaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
Doing wat + 1 results in wat1wat1wat1wat1...
I suspect this is just straightforward behaviour that trying to subtract a number from a string results in NaN.
Here's a list of explanations for the results you're seeing (and supposed to be seeing). The references I'm using are from the ECMA-262 standard.
[] + []
When using the addition operator, both the left and right operands are converted to primitives first (§11.6.1). As per §9.1, converting an object (in this case an array) to a primitive returns its default value, which for objects with a valid toString() method is the result of calling object.toString() (§8.12.8). For arrays this is the same as calling array.join() (§15.4.4.2). Joining an empty array results in an empty string, so step #7 of the addition operator returns the concatenation of two empty strings, which is the empty string.
[] + {}
Similar to [] + [], both operands are converted to primitives first. For "Object objects" (§15.2), this is again the result of calling object.toString(), which for non-null, non-undefined objects is "[object Object]" (§15.2.4.2).
{} + []
The {} here is not parsed as an object, but instead as an empty block (§12.1, at least as long as you're not forcing that statement to be an expression, but more about that later). The return value of empty blocks is empty, so the result of that statement is the same as +[]. The unary + operator (§11.4.6) returns ToNumber(ToPrimitive(operand)). As we already know, ToPrimitive([]) is the empty string, and according to §9.3.1, ToNumber("") is 0.
{} + {}
Similar to the previous case, the first {} is parsed as a block with empty return value. Again, +{} is the same as ToNumber(ToPrimitive({})), and ToPrimitive({}) is "[object Object]" (see [] + {}). So to get the result of +{}, we have to apply ToNumber on the string "[object Object]". When following the steps from §9.3.1, we get NaN as a result:
If the grammar cannot interpret the String as an expansion of StringNumericLiteral, then the result of ToNumber is NaN.
Array(16).join("wat" - 1)
As per §15.4.1.1 and §15.4.2.2, Array(16) creates a new array with length 16. To get the value of the argument to join, §11.6.2 steps #5 and #6 show that we have to convert both operands to a number using ToNumber. ToNumber(1) is simply 1 (§9.3), whereas ToNumber("wat") again is NaN as per §9.3.1. Following step 7 of §11.6.2, §11.6.3 dictates that
If either operand is NaN, the result is NaN.
So the argument to Array(16).join is NaN. Following §15.4.4.5 (Array.prototype.join), we have to call ToString on the argument, which is "NaN" (§9.8.1):
If m is NaN, return the String "NaN".
Following step 10 of §15.4.4.5, we get 15 repetitions of the concatenation of "NaN" and the empty string, which equals the result you're seeing.
When using "wat" + 1 instead of "wat" - 1 as argument, the addition operator converts 1 to a string instead of converting "wat" to a number, so it effectively calls Array(16).join("wat1").
As to why you're seeing different results for the {} + [] case: When using it as a function argument, you're forcing the statement to be an ExpressionStatement, which makes it impossible to parse {} as empty block, so it's instead parsed as an empty object literal.
This is more of a comment than an answer, but for some reason I can't comment on your question. I wanted to correct your JSFiddle code. However, I posted this on Hacker News and someone suggested that I repost it here.
The problem in the JSFiddle code is that ({}) (opening braces inside of parentheses) is not the same as {} (opening braces as the start of a line of code). So when you type out({} + []) you are forcing the {} to be something which it is not when you type {} + []. This is part of the overall 'wat'-ness of Javascript.
The basic idea was simple JavaScript wanted to allow both of these forms:
if (u)
v;
if (x) {
y;
z;
}
To do so, two interpretations were made of the opening brace: 1. it is not required and 2. it can appear anywhere.
This was a wrong move. Real code doesn't have an opening brace appearing in the middle of nowhere, and real code also tends to be more fragile when it uses the first form rather than the second. (About once every other month at my last job, I'd get called to a coworker's desk when their modifications to my code weren't working, and the problem was that they'd added a line to the "if" without adding curly braces. I eventually just adopted the habit that the curly braces are always required, even when you're only writing one line.)
Fortunately in many cases eval() will replicate the full wat-ness of JavaScript. The JSFiddle code should read:
function out(code) {
function format(x) {
return typeof x === "string" ?
JSON.stringify(x) : x;
}
document.writeln('>>> ' + code);
document.writeln(format(eval(code)));
}
document.writeln("<pre>");
out('[] + []');
out('[] + {}');
out('{} + []');
out('{} + {}');
out('Array(16).join("wat" + 1)');
out('Array(16).join("wat - 1")');
out('Array(16).join("wat" - 1) + " Batman!"');
document.writeln("</pre>");
[Also that is the first time I have written document.writeln in many many many years, and I feel a little dirty writing anything involving both document.writeln() and eval().]
I second #Ventero’s solution. If you want to, you can go into more detail as to how + converts its operands.
First step (§9.1): convert both operands to primitives (primitive values are undefined, null, booleans, numbers, strings; all other values are objects, including arrays and functions). If an operand is already primitive, you are done. If not, it is an object obj and the following steps are performed:
Call obj.valueOf(). If it returns a primitive, you are done. Direct instances of Object and arrays return themselves, so you are not done yet.
Call obj.toString(). If it returns a primitive, you are done. {} and [] both return a string, so you are done.
Otherwise, throw a TypeError.
For dates, step 1 and 2 are swapped. You can observe the conversion behavior as follows:
var obj = {
valueOf: function () {
console.log("valueOf");
return {}; // not a primitive
},
toString: function () {
console.log("toString");
return {}; // not a primitive
}
}
Interaction (Number() first converts to primitive then to number):
> Number(obj)
valueOf
toString
TypeError: Cannot convert object to primitive value
Second step (§11.6.1): If one of the operands is a string, the other operand is also converted to string and the result is produced by concatenating two strings. Otherwise, both operands are converted to numbers and the result is produced by adding them.
More detailed explanation of the conversion process: “What is {} + {} in JavaScript?”
We may refer to the specification and that's great and most accurate, but most of the cases can also be explained in a more comprehensible way with the following statements:
+ and - operators work only with primitive values. More specifically +(addition) works with either strings or numbers, and +(unary) and -(subtraction and unary) works only with numbers.
All native functions or operators that expect primitive value as argument, will first convert that argument to desired primitive type. It is done with valueOf or toString, which are available on any object. That's the reason why such functions or operators don't throw errors when invoked on objects.
So we may say that:
[] + [] is same as String([]) + String([]) which is same as '' + ''. I mentioned above that +(addition) is also valid for numbers, but there is no valid number representation of an array in JavaScript, so addition of strings is used instead.
[] + {} is same as String([]) + String({}) which is same as '' + '[object Object]'
{} + []. This one deserves more explanation (see Ventero answer). In that case, curly braces are treated not as an object but as an empty block, so it turns out to be same as +[]. Unary + works only with numbers, so the implementation tries to get a number out of []. First it tries valueOf which in the case of arrays returns the same object, so then it tries the last resort: conversion of a toString result to a number. We may write it as +Number(String([])) which is same as +Number('') which is same as +0.
Array(16).join("wat" - 1) subtraction - works only with numbers, so it's the same as: Array(16).join(Number("wat") - 1), as "wat" can't be converted to a valid number. We receive NaN, and any arithmetic operation on NaN results with NaN, so we have: Array(16).join(NaN).
To buttress what has been shared earlier.
The underlying cause of this behaviour is partly due to the weakly-typed nature of JavaScript. For example, the expression 1 + “2” is ambiguous since there are two possible interpretations based on the operand types (int, string) and (int int):
User intends to concatenate two strings, result: “12”
User intends to add two numbers, result: 3
Thus with varying input types,the output possibilities increase.
The addition algorithm
Coerce operands to primitive values
The JavaScript primitives are string, number, null, undefined and boolean (Symbol is coming soon in ES6). Any other value is an object (e.g. arrays, functions and objects). The coercion process for converting objects into primitive values is described thus:
If a primitive value is returned when object.valueOf() is invoked, then return this value, otherwise continue
If a primitive value is returned when object.toString() is invoked, then return this value, otherwise continue
Throw a TypeError
Note: For date values, the order is to invoke toString before valueOf.
If any operand value is a string, then do a string concatenation
Otherwise, convert both operands to their numeric value and then add these values
Knowing the various coercion values of types in JavaScript does help to make the confusing outputs clearer. See the coercion table below
+-----------------+-------------------+---------------+
| Primitive Value | String value | Numeric value |
+-----------------+-------------------+---------------+
| null | “null” | 0 |
| undefined | “undefined” | NaN |
| true | “true” | 1 |
| false | “false” | 0 |
| 123 | “123” | 123 |
| [] | “” | 0 |
| {} | “[object Object]” | NaN |
+-----------------+-------------------+---------------+
It is also good to know that JavaScript's + operator is left-associative as this determines what the output will be cases involving more than one + operation.
Leveraging the
Thus 1 + "2" will give "12" because any addition involving a string will always default to string concatenation.
You can read more examples in this blog post (disclaimer I wrote it).

About "5"-1 & "5"+1 in Javascript (plus and minus signs) [duplicate]

This question already has answers here:
Why does JavaScript handle the plus and minus operators between strings and numbers differently?
(7 answers)
Closed 7 years ago.
I read a book about operators in Javascript, and this confused me.
console.log("5"+1);
This would make "5" as a string. So the result would be 51.
console.log("5"-1);
This result would be 4. I know it converts "5" to 5, but why it isn't shown undefined as "a string minus a number"?
Update: So how about other languages? Are they more restrict?
Sadly, it was expected from JavaScript to ride on Java's success for promotion on its early days and the plus for string concatenation was adopted since Java used it.
So JavaScript tries hard to coerce strings into numbers for you, it really does, its just that the plus was taken for strings so....well...
While Javascript has many strenghts it was made in 10 days and has many hilarious aspects like this one, check this comedy gold
The + is a operator that means SUM when adding numbers and that means CONCATENATE when using Strings.
As the first is a STRING, it will continue concatenating a "5"+toString(1).
As the MINUS (-) operator does not work with String you are getting undefined.
If you want to use MINUS operator, you will need to do :
parseInt("5") -> It will give you 5, the number
parseInt("5")-1 = 4
"5"+1 = 51
parseInt("5")+1 = 6
Hope it will help you
Because when we use '+' it can be used in two different ways:-
1. as mathematical operator.
2. to concatenate strings
but '-' can only be used as mathematical operator.
Hence javascript considers '5' as numerics in case of '-' while '5' as string in case of '+'.
In javascript (+) operator operates the way described below
3+true will return 4 , (+) operator between a number and a boolean or two boolean will convert boolean to number , hence true is converted to 1 hence the result is 4
"2"+true will return "2true" , if one of the operand is string it will convert the other operand (number or boolean) to string and process the concatenation
-"12"+3 will return -9 , (-) operator in front of string will convert the string to number and will make it as -12 and return -9
According to the standard EcmaScript 262. The + and - operators behave differently when strings are involved. The first converts every value to a string. The second converts every value to a number.
From the standard:
If Type(lprim) is String or Type(rprim) is String, then Return the
String that is the result of concatenating ToString(lprim) followed by
ToString(rprim)
This rules implies that if in the expression there is a string value, all values involved in the + operation are converted to a string. In JavaScript when the + operator is used with strings, it concatenates them. This is why console.log("5"+1) returns "51". 1 is converted to a string and then, "5" + "1" are concatenated together.
Nevertheless, the above rule doesn't apply for the - operator. When you are using a - all values are converted to numbers according to the Standard (see below). Therefore, in this case, "5" is converted to 5 and then 1 is subtracted.
From the standard:
5 Let lnum be ToNumber(lval).
6 Let rnum be ToNumber(rval).
Operator definition from the standard EcmaScript 262.
Operator + : http://www.ecma-international.org/ecma-262/5.1/#sec-11.6.1
Operator - : http://www.ecma-international.org/ecma-262/5.1/#sec-11.6.2
Because of the type coercion and how it isn't very consistent in JavaScript, in the second case the "5" is converted to a number 5, and 1 is subtracted from it, giving you 4.
"5" could be coerced to 5 (Integer). That's why you get 4 as output.
However if you try:
console.log("text" - 1);
Text cannot be coerced, and the output is NaN
The subtraction operator (-) subtracts the number to the right of the operator from the number on the left.
When either of the operands are strings, an attempt is made to convert the strings to numbers.
Instead of using "5" if you try console.log("abc" - 1); it will prompt a error as NaN.
Just for the info:
The subtract operator has special rules to deal with the variety of type conversions present in JavaScript:
If the two operands are numbers, perform arithmetic subtract and return the result.
If either number is NaN, the result is NaN.
If Infinity is subtracted from Infinity, the result is NaN.
If –Infinity is subtracted from –Infinity, the result is NaN.
If –Infinity is subtracted from Infinity, the result is Infinity.
If Infinity is subtracted from –Infinity, the result is –Infinity.
If +0 is subtracted from +0, the result is +0.
If –0 is subtracted from +0, the result is –0.
If –0 is subtracted from –0, the result is +0.
If either of the two operands is not a number, the result is NaN.

How [1,2] + [4,5,6][1] = 1,25 in JavaScript [duplicate]

This question already has answers here:
Why is [1,2] + [3,4] = "1,23,4" in JavaScript?
(14 answers)
Closed 7 years ago.
I got this question from Interview,
[1,2] + [4,5,6][1]
JavaScript giving answer 1,25.
How it's happening? Please explain clearly.
Lets start with the last part and write it more verbose
var arr = [4,5,6];
var value = arr[1];
Is the same as
[4,5,6][1]
and as arrays are zero based, that gives 5, so we really have
[1,2] + 5;
The first part is equal to
[1,2].toString(),
and that returns the string "1,2" because the array is converted to a string when trying to use it in an expression like that, as arrays can't be added to numbers, and then we have
"1,2" + 5
Concatenating strings with numbers gives strings, and adding the strings together we get
"1,25"
It breaks down like this:
Starting with:
[1,2] + [4,5,6][1]
First, each side gets evaluated, and since the right-hand side is an array initializer and lookup, it comes out to 5:
[1,2] + 5
Now the + operator starts its work. It isn't defined for arrays, the first thing it does is try to convert its operands into either strings or numbers. In the case of an array, it'll be a string as though from Array#toString, which does Array#join, giving us:
"1,2" + 5
When you use + where either side is a string, the result is string concatenation.
First, [4,5,6][1] evaluates to the number 5
Then, the + operator is being applied to a first argument which is an Array and not a Number, so javascript assumes you're doing a string concatenation, not addition. Your array [1,2] becomes a string, which is "1,2". You can see this yourself with [1,2].toString().
The number 5 is now being appended to a string, so it to gets converted to a string, and appended together to get "1,25".

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