Related
I am debugging some JavaScript and can't explain what this || does:
function (title, msg) {
var title = title || 'Error';
var msg = msg || 'Error on Request';
}
Why is this guy using var title = title || 'ERROR'? I sometimes see it without a var declaration as well.
What is the double pipe operator (||)?
The double pipe operator (||) is the logical OR operator . In most languages it works the following way:
If the first value is false, it checks the second value. If that's true, it returns true and if the second value is false, it returns false.
If the first value is true, it always returns true, no matter what the second value is.
So basically it works like this function:
function or(x, y) {
if (x) {
return true;
} else if (y) {
return true;
} else {
return false;
}
}
If you still don't understand, look at this table:
| true false
------+---------------
true | true true
false | true false
In other words, it's only false when both values are false.
How is it different in JavaScript?
JavaScript is a bit different, because it's a loosely typed language. In this case it means that you can use || operator with values that are not booleans. Though it makes no sense, you can use this operator with for example a function and an object:
(function(){}) || {}
What happens there?
If values are not boolean, JavaScript makes implicit conversion to boolean. It means that if the value is falsey (e.g. 0, "", null, undefined (see also All falsey values in JavaScript)), it will be treated as false; otherwise it's treated as true.
So the above example should give true, because empty function is truthy. Well, it doesn't. It returns the empty function. That's because JavaScript's || operator doesn't work as I wrote at the beginning. It works the following way:
If the first value is falsey, it returns the second value.
If the first value is truthy, it returns the first value.
Surprised? Actually, it's "compatible" with the traditional || operator. It could be written as following function:
function or(x, y) {
if (x) {
return x;
} else {
return y;
}
}
If you pass a truthy value as x, it returns x, that is, a truthy value. So if you use it later in if clause:
(function(x, y) {
var eitherXorY = x || y;
if (eitherXorY) {
console.log("Either x or y is truthy.");
} else {
console.log("Neither x nor y is truthy");
}
}(true/*, undefined*/));
you get "Either x or y is truthy.".
If x was falsey, eitherXorY would be y. In this case you would get the "Either x or y is truthy." if y was truthy; otherwise you'd get "Neither x nor y is truthy".
The actual question
Now, when you know how || operator works, you can probably make out by yourself what does x = x || y mean. If x is truthy, x is assigned to x, so actually nothing happens; otherwise y is assigned to x. It is commonly used to define default parameters in functions. However, it is often considered a bad programming practice, because it prevents you from passing a falsey value (which is not necessarily undefined or null) as a parameter. Consider following example:
function badFunction(/* boolean */flagA) {
flagA = flagA || true;
console.log("flagA is set to " + (flagA ? "true" : "false"));
}
It looks valid at the first sight. However, what would happen if you passed false as flagA parameter (since it's boolean, i.e. can be true or false)? It would become true. In this example, there is no way to set flagA to false.
It would be a better idea to explicitly check whether flagA is undefined, like that:
function goodFunction(/* boolean */flagA) {
flagA = typeof flagA !== "undefined" ? flagA : true;
console.log("flagA is set to " + (flagA ? "true" : "false"));
}
Though it's longer, it always works and it's easier to understand.
You can also use the ES6 syntax for default function parameters, but note that it doesn't work in older browsers (like IE). If you want to support these browsers, you should transpile your code with Babel.
See also Logical Operators on MDN.
It means the title argument is optional. So if you call the method with no arguments it will use a default value of "Error".
It's shorthand for writing:
if (!title) {
title = "Error";
}
This kind of shorthand trick with boolean expressions is common in Perl too. With the expression:
a OR b
it evaluates to true if either a or b is true. So if a is true you don't need to check b at all. This is called short-circuit boolean evaluation so:
var title = title || "Error";
basically checks if title evaluates to false. If it does, it "returns" "Error", otherwise it returns title.
If title is not set, use 'ERROR' as default value.
More generic:
var foobar = foo || default;
Reads: Set foobar to foo or default.
You could even chain this up many times:
var foobar = foo || bar || something || 42;
Explaining this a little more...
The || operator is the logical-or operator. The result is true if the first part is true and it is true if the second part is true and it is true if both parts are true. For clarity, here it is in a table:
X | Y | X || Y
---+---+--------
F | F | F
---+---+--------
F | T | T
---+---+--------
T | F | T
---+---+--------
T | T | T
---+---+--------
Now notice something here? If X is true, the result is always true. So if we know that X is true we don't have to check Y at all. Many languages thus implement "short circuit" evaluators for logical-or (and logical-and coming from the other direction). They check the first element and if that's true they don't bother checking the second at all. The result (in logical terms) is the same, but in terms of execution there's potentially a huge difference if the second element is expensive to calculate.
So what does this have to do with your example?
var title = title || 'Error';
Let's look at that. The title element is passed in to your function. In JavaScript if you don't pass in a parameter, it defaults to a null value. Also in JavaScript if your variable is a null value it is considered to be false by the logical operators. So if this function is called with a title given, it is a non-false value and thus assigned to the local variable. If, however, it is not given a value, it is a null value and thus false. The logical-or operator then evaluates the second expression and returns 'Error' instead. So now the local variable is given the value 'Error'.
This works because of the implementation of logical expressions in JavaScript. It doesn't return a proper boolean value (true or false) but instead returns the value it was given under some rules as to what's considered equivalent to true and what's considered equivalent to false. Look up your JavaScript reference to learn what JavaScript considers to be true or false in boolean contexts.
|| is the boolean OR operator. As in JavaScript, undefined, null, 0, false are considered as falsy values.
It simply means
true || true = true
false || true = true
true || false = true
false || false = false
undefined || "value" = "value"
"value" || undefined = "value"
null || "value" = "value"
"value" || null = "value"
0 || "value" = "value"
"value" || 0 = "value"
false || "value" = "value"
"value" || false = "value"
Basically, it checks if the value before the || evaluates to true. If yes, it takes this value, and if not, it takes the value after the ||.
Values for which it will take the value after the || (as far as I remember):
undefined
false
0
'' (Null or Null string)
Whilst Cletus' answer is correct, I feel more detail should be added in regards to "evaluates to false" in JavaScript.
var title = title || 'Error';
var msg = msg || 'Error on Request';
Is not just checking if title/msg has been provided, but also if either of them are falsy. i.e. one of the following:
false.
0 (zero)
"" (empty string)
null.
undefined.
NaN (a special Number value meaning Not-a-Number!)
So in the line
var title = title || 'Error';
If title is truthy (i.e., not falsy, so title = "titleMessage" etc.) then the Boolean OR (||) operator has found one 'true' value, which means it evaluates to true, so it short-circuits and returns the true value (title).
If title is falsy (i.e. one of the list above), then the Boolean OR (||) operator has found a 'false' value, and now needs to evaluate the other part of the operator, 'Error', which evaluates to true, and is hence returned.
It would also seem (after some quick firebug console experimentation) if both sides of the operator evaluate to false, it returns the second 'falsy' operator.
i.e.
return ("" || undefined)
returns undefined, this is probably to allow you to use the behavior asked about in this question when trying to default title/message to "". i.e. after running
var foo = undefined
foo = foo || ""
foo would be set to ""
Double pipe stands for logical "OR". This is not really the case when the "parameter not set", since strictly in JavaScript if you have code like this:
function foo(par) {
}
Then calls
foo()
foo("")
foo(null)
foo(undefined)
foo(0)
are not equivalent.
Double pipe (||) will cast the first argument to Boolean and if the resulting Boolean is true - do the assignment, otherwise it will assign the right part.
This matters if you check for unset parameter.
Let's say, we have a function setSalary that has one optional parameter. If the user does not supply the parameter then the default value of 10 should be used.
If you do the check like this:
function setSalary(dollars) {
salary = dollars || 10
}
This will give an unexpected result for a call like:
setSalary(0)
It will still set the 10 following the flow described above.
Double pipe operator
This example may be useful:
var section = document.getElementById('special');
if(!section){
section = document.getElementById('main');
}
It can also be:
var section = document.getElementById('special') || document.getElementById('main');
To add some explanation to all said before me, I should give you some examples to understand logical concepts.
var name = false || "Mohsen"; # name equals to Mohsen
var family = true || "Alizadeh" # family equals to true
It means if the left side evaluated as a true statement it will be finished and the left side will be returned and assigned to the variable. in other cases the right side will be returned and assigned.
And operator have the opposite structure like below.
var name = false && "Mohsen" # name equals to false
var family = true && "Alizadeh" # family equals to Alizadeh
Quote: "What does the construct x = x || y mean?"
Assigning a default value.
This means providing a default value of y to x,
in case x is still waiting for its value but hasn't received it yet or was deliberately omitted in order to fall back to a default.
And I have to add one more thing: This bit of shorthand is an abomination. It misuses an accidental interpreter optimization (not bothering with the second operation if the first is truthy) to control an assignment. That use has nothing to do with the purpose of the operator. I do not believe it should ever be used.
I prefer the ternary operator for initialization, for example,
var title = title?title:'Error';
This uses a one-line conditional operation for its correct purpose. It still plays unsightly games with truthiness but, that's JavaScript for you.
I am debugging some JavaScript and can't explain what this || does:
function (title, msg) {
var title = title || 'Error';
var msg = msg || 'Error on Request';
}
Why is this guy using var title = title || 'ERROR'? I sometimes see it without a var declaration as well.
What is the double pipe operator (||)?
The double pipe operator (||) is the logical OR operator . In most languages it works the following way:
If the first value is false, it checks the second value. If that's true, it returns true and if the second value is false, it returns false.
If the first value is true, it always returns true, no matter what the second value is.
So basically it works like this function:
function or(x, y) {
if (x) {
return true;
} else if (y) {
return true;
} else {
return false;
}
}
If you still don't understand, look at this table:
| true false
------+---------------
true | true true
false | true false
In other words, it's only false when both values are false.
How is it different in JavaScript?
JavaScript is a bit different, because it's a loosely typed language. In this case it means that you can use || operator with values that are not booleans. Though it makes no sense, you can use this operator with for example a function and an object:
(function(){}) || {}
What happens there?
If values are not boolean, JavaScript makes implicit conversion to boolean. It means that if the value is falsey (e.g. 0, "", null, undefined (see also All falsey values in JavaScript)), it will be treated as false; otherwise it's treated as true.
So the above example should give true, because empty function is truthy. Well, it doesn't. It returns the empty function. That's because JavaScript's || operator doesn't work as I wrote at the beginning. It works the following way:
If the first value is falsey, it returns the second value.
If the first value is truthy, it returns the first value.
Surprised? Actually, it's "compatible" with the traditional || operator. It could be written as following function:
function or(x, y) {
if (x) {
return x;
} else {
return y;
}
}
If you pass a truthy value as x, it returns x, that is, a truthy value. So if you use it later in if clause:
(function(x, y) {
var eitherXorY = x || y;
if (eitherXorY) {
console.log("Either x or y is truthy.");
} else {
console.log("Neither x nor y is truthy");
}
}(true/*, undefined*/));
you get "Either x or y is truthy.".
If x was falsey, eitherXorY would be y. In this case you would get the "Either x or y is truthy." if y was truthy; otherwise you'd get "Neither x nor y is truthy".
The actual question
Now, when you know how || operator works, you can probably make out by yourself what does x = x || y mean. If x is truthy, x is assigned to x, so actually nothing happens; otherwise y is assigned to x. It is commonly used to define default parameters in functions. However, it is often considered a bad programming practice, because it prevents you from passing a falsey value (which is not necessarily undefined or null) as a parameter. Consider following example:
function badFunction(/* boolean */flagA) {
flagA = flagA || true;
console.log("flagA is set to " + (flagA ? "true" : "false"));
}
It looks valid at the first sight. However, what would happen if you passed false as flagA parameter (since it's boolean, i.e. can be true or false)? It would become true. In this example, there is no way to set flagA to false.
It would be a better idea to explicitly check whether flagA is undefined, like that:
function goodFunction(/* boolean */flagA) {
flagA = typeof flagA !== "undefined" ? flagA : true;
console.log("flagA is set to " + (flagA ? "true" : "false"));
}
Though it's longer, it always works and it's easier to understand.
You can also use the ES6 syntax for default function parameters, but note that it doesn't work in older browsers (like IE). If you want to support these browsers, you should transpile your code with Babel.
See also Logical Operators on MDN.
It means the title argument is optional. So if you call the method with no arguments it will use a default value of "Error".
It's shorthand for writing:
if (!title) {
title = "Error";
}
This kind of shorthand trick with boolean expressions is common in Perl too. With the expression:
a OR b
it evaluates to true if either a or b is true. So if a is true you don't need to check b at all. This is called short-circuit boolean evaluation so:
var title = title || "Error";
basically checks if title evaluates to false. If it does, it "returns" "Error", otherwise it returns title.
If title is not set, use 'ERROR' as default value.
More generic:
var foobar = foo || default;
Reads: Set foobar to foo or default.
You could even chain this up many times:
var foobar = foo || bar || something || 42;
Explaining this a little more...
The || operator is the logical-or operator. The result is true if the first part is true and it is true if the second part is true and it is true if both parts are true. For clarity, here it is in a table:
X | Y | X || Y
---+---+--------
F | F | F
---+---+--------
F | T | T
---+---+--------
T | F | T
---+---+--------
T | T | T
---+---+--------
Now notice something here? If X is true, the result is always true. So if we know that X is true we don't have to check Y at all. Many languages thus implement "short circuit" evaluators for logical-or (and logical-and coming from the other direction). They check the first element and if that's true they don't bother checking the second at all. The result (in logical terms) is the same, but in terms of execution there's potentially a huge difference if the second element is expensive to calculate.
So what does this have to do with your example?
var title = title || 'Error';
Let's look at that. The title element is passed in to your function. In JavaScript if you don't pass in a parameter, it defaults to a null value. Also in JavaScript if your variable is a null value it is considered to be false by the logical operators. So if this function is called with a title given, it is a non-false value and thus assigned to the local variable. If, however, it is not given a value, it is a null value and thus false. The logical-or operator then evaluates the second expression and returns 'Error' instead. So now the local variable is given the value 'Error'.
This works because of the implementation of logical expressions in JavaScript. It doesn't return a proper boolean value (true or false) but instead returns the value it was given under some rules as to what's considered equivalent to true and what's considered equivalent to false. Look up your JavaScript reference to learn what JavaScript considers to be true or false in boolean contexts.
|| is the boolean OR operator. As in JavaScript, undefined, null, 0, false are considered as falsy values.
It simply means
true || true = true
false || true = true
true || false = true
false || false = false
undefined || "value" = "value"
"value" || undefined = "value"
null || "value" = "value"
"value" || null = "value"
0 || "value" = "value"
"value" || 0 = "value"
false || "value" = "value"
"value" || false = "value"
Basically, it checks if the value before the || evaluates to true. If yes, it takes this value, and if not, it takes the value after the ||.
Values for which it will take the value after the || (as far as I remember):
undefined
false
0
'' (Null or Null string)
Whilst Cletus' answer is correct, I feel more detail should be added in regards to "evaluates to false" in JavaScript.
var title = title || 'Error';
var msg = msg || 'Error on Request';
Is not just checking if title/msg has been provided, but also if either of them are falsy. i.e. one of the following:
false.
0 (zero)
"" (empty string)
null.
undefined.
NaN (a special Number value meaning Not-a-Number!)
So in the line
var title = title || 'Error';
If title is truthy (i.e., not falsy, so title = "titleMessage" etc.) then the Boolean OR (||) operator has found one 'true' value, which means it evaluates to true, so it short-circuits and returns the true value (title).
If title is falsy (i.e. one of the list above), then the Boolean OR (||) operator has found a 'false' value, and now needs to evaluate the other part of the operator, 'Error', which evaluates to true, and is hence returned.
It would also seem (after some quick firebug console experimentation) if both sides of the operator evaluate to false, it returns the second 'falsy' operator.
i.e.
return ("" || undefined)
returns undefined, this is probably to allow you to use the behavior asked about in this question when trying to default title/message to "". i.e. after running
var foo = undefined
foo = foo || ""
foo would be set to ""
Double pipe stands for logical "OR". This is not really the case when the "parameter not set", since strictly in JavaScript if you have code like this:
function foo(par) {
}
Then calls
foo()
foo("")
foo(null)
foo(undefined)
foo(0)
are not equivalent.
Double pipe (||) will cast the first argument to Boolean and if the resulting Boolean is true - do the assignment, otherwise it will assign the right part.
This matters if you check for unset parameter.
Let's say, we have a function setSalary that has one optional parameter. If the user does not supply the parameter then the default value of 10 should be used.
If you do the check like this:
function setSalary(dollars) {
salary = dollars || 10
}
This will give an unexpected result for a call like:
setSalary(0)
It will still set the 10 following the flow described above.
Double pipe operator
This example may be useful:
var section = document.getElementById('special');
if(!section){
section = document.getElementById('main');
}
It can also be:
var section = document.getElementById('special') || document.getElementById('main');
To add some explanation to all said before me, I should give you some examples to understand logical concepts.
var name = false || "Mohsen"; # name equals to Mohsen
var family = true || "Alizadeh" # family equals to true
It means if the left side evaluated as a true statement it will be finished and the left side will be returned and assigned to the variable. in other cases the right side will be returned and assigned.
And operator have the opposite structure like below.
var name = false && "Mohsen" # name equals to false
var family = true && "Alizadeh" # family equals to Alizadeh
Quote: "What does the construct x = x || y mean?"
Assigning a default value.
This means providing a default value of y to x,
in case x is still waiting for its value but hasn't received it yet or was deliberately omitted in order to fall back to a default.
And I have to add one more thing: This bit of shorthand is an abomination. It misuses an accidental interpreter optimization (not bothering with the second operation if the first is truthy) to control an assignment. That use has nothing to do with the purpose of the operator. I do not believe it should ever be used.
I prefer the ternary operator for initialization, for example,
var title = title?title:'Error';
This uses a one-line conditional operation for its correct purpose. It still plays unsightly games with truthiness but, that's JavaScript for you.
I am debugging some JavaScript and can't explain what this || does:
function (title, msg) {
var title = title || 'Error';
var msg = msg || 'Error on Request';
}
Why is this guy using var title = title || 'ERROR'? I sometimes see it without a var declaration as well.
What is the double pipe operator (||)?
The double pipe operator (||) is the logical OR operator . In most languages it works the following way:
If the first value is false, it checks the second value. If that's true, it returns true and if the second value is false, it returns false.
If the first value is true, it always returns true, no matter what the second value is.
So basically it works like this function:
function or(x, y) {
if (x) {
return true;
} else if (y) {
return true;
} else {
return false;
}
}
If you still don't understand, look at this table:
| true false
------+---------------
true | true true
false | true false
In other words, it's only false when both values are false.
How is it different in JavaScript?
JavaScript is a bit different, because it's a loosely typed language. In this case it means that you can use || operator with values that are not booleans. Though it makes no sense, you can use this operator with for example a function and an object:
(function(){}) || {}
What happens there?
If values are not boolean, JavaScript makes implicit conversion to boolean. It means that if the value is falsey (e.g. 0, "", null, undefined (see also All falsey values in JavaScript)), it will be treated as false; otherwise it's treated as true.
So the above example should give true, because empty function is truthy. Well, it doesn't. It returns the empty function. That's because JavaScript's || operator doesn't work as I wrote at the beginning. It works the following way:
If the first value is falsey, it returns the second value.
If the first value is truthy, it returns the first value.
Surprised? Actually, it's "compatible" with the traditional || operator. It could be written as following function:
function or(x, y) {
if (x) {
return x;
} else {
return y;
}
}
If you pass a truthy value as x, it returns x, that is, a truthy value. So if you use it later in if clause:
(function(x, y) {
var eitherXorY = x || y;
if (eitherXorY) {
console.log("Either x or y is truthy.");
} else {
console.log("Neither x nor y is truthy");
}
}(true/*, undefined*/));
you get "Either x or y is truthy.".
If x was falsey, eitherXorY would be y. In this case you would get the "Either x or y is truthy." if y was truthy; otherwise you'd get "Neither x nor y is truthy".
The actual question
Now, when you know how || operator works, you can probably make out by yourself what does x = x || y mean. If x is truthy, x is assigned to x, so actually nothing happens; otherwise y is assigned to x. It is commonly used to define default parameters in functions. However, it is often considered a bad programming practice, because it prevents you from passing a falsey value (which is not necessarily undefined or null) as a parameter. Consider following example:
function badFunction(/* boolean */flagA) {
flagA = flagA || true;
console.log("flagA is set to " + (flagA ? "true" : "false"));
}
It looks valid at the first sight. However, what would happen if you passed false as flagA parameter (since it's boolean, i.e. can be true or false)? It would become true. In this example, there is no way to set flagA to false.
It would be a better idea to explicitly check whether flagA is undefined, like that:
function goodFunction(/* boolean */flagA) {
flagA = typeof flagA !== "undefined" ? flagA : true;
console.log("flagA is set to " + (flagA ? "true" : "false"));
}
Though it's longer, it always works and it's easier to understand.
You can also use the ES6 syntax for default function parameters, but note that it doesn't work in older browsers (like IE). If you want to support these browsers, you should transpile your code with Babel.
See also Logical Operators on MDN.
It means the title argument is optional. So if you call the method with no arguments it will use a default value of "Error".
It's shorthand for writing:
if (!title) {
title = "Error";
}
This kind of shorthand trick with boolean expressions is common in Perl too. With the expression:
a OR b
it evaluates to true if either a or b is true. So if a is true you don't need to check b at all. This is called short-circuit boolean evaluation so:
var title = title || "Error";
basically checks if title evaluates to false. If it does, it "returns" "Error", otherwise it returns title.
If title is not set, use 'ERROR' as default value.
More generic:
var foobar = foo || default;
Reads: Set foobar to foo or default.
You could even chain this up many times:
var foobar = foo || bar || something || 42;
Explaining this a little more...
The || operator is the logical-or operator. The result is true if the first part is true and it is true if the second part is true and it is true if both parts are true. For clarity, here it is in a table:
X | Y | X || Y
---+---+--------
F | F | F
---+---+--------
F | T | T
---+---+--------
T | F | T
---+---+--------
T | T | T
---+---+--------
Now notice something here? If X is true, the result is always true. So if we know that X is true we don't have to check Y at all. Many languages thus implement "short circuit" evaluators for logical-or (and logical-and coming from the other direction). They check the first element and if that's true they don't bother checking the second at all. The result (in logical terms) is the same, but in terms of execution there's potentially a huge difference if the second element is expensive to calculate.
So what does this have to do with your example?
var title = title || 'Error';
Let's look at that. The title element is passed in to your function. In JavaScript if you don't pass in a parameter, it defaults to a null value. Also in JavaScript if your variable is a null value it is considered to be false by the logical operators. So if this function is called with a title given, it is a non-false value and thus assigned to the local variable. If, however, it is not given a value, it is a null value and thus false. The logical-or operator then evaluates the second expression and returns 'Error' instead. So now the local variable is given the value 'Error'.
This works because of the implementation of logical expressions in JavaScript. It doesn't return a proper boolean value (true or false) but instead returns the value it was given under some rules as to what's considered equivalent to true and what's considered equivalent to false. Look up your JavaScript reference to learn what JavaScript considers to be true or false in boolean contexts.
|| is the boolean OR operator. As in JavaScript, undefined, null, 0, false are considered as falsy values.
It simply means
true || true = true
false || true = true
true || false = true
false || false = false
undefined || "value" = "value"
"value" || undefined = "value"
null || "value" = "value"
"value" || null = "value"
0 || "value" = "value"
"value" || 0 = "value"
false || "value" = "value"
"value" || false = "value"
Basically, it checks if the value before the || evaluates to true. If yes, it takes this value, and if not, it takes the value after the ||.
Values for which it will take the value after the || (as far as I remember):
undefined
false
0
'' (Null or Null string)
Whilst Cletus' answer is correct, I feel more detail should be added in regards to "evaluates to false" in JavaScript.
var title = title || 'Error';
var msg = msg || 'Error on Request';
Is not just checking if title/msg has been provided, but also if either of them are falsy. i.e. one of the following:
false.
0 (zero)
"" (empty string)
null.
undefined.
NaN (a special Number value meaning Not-a-Number!)
So in the line
var title = title || 'Error';
If title is truthy (i.e., not falsy, so title = "titleMessage" etc.) then the Boolean OR (||) operator has found one 'true' value, which means it evaluates to true, so it short-circuits and returns the true value (title).
If title is falsy (i.e. one of the list above), then the Boolean OR (||) operator has found a 'false' value, and now needs to evaluate the other part of the operator, 'Error', which evaluates to true, and is hence returned.
It would also seem (after some quick firebug console experimentation) if both sides of the operator evaluate to false, it returns the second 'falsy' operator.
i.e.
return ("" || undefined)
returns undefined, this is probably to allow you to use the behavior asked about in this question when trying to default title/message to "". i.e. after running
var foo = undefined
foo = foo || ""
foo would be set to ""
Double pipe stands for logical "OR". This is not really the case when the "parameter not set", since strictly in JavaScript if you have code like this:
function foo(par) {
}
Then calls
foo()
foo("")
foo(null)
foo(undefined)
foo(0)
are not equivalent.
Double pipe (||) will cast the first argument to Boolean and if the resulting Boolean is true - do the assignment, otherwise it will assign the right part.
This matters if you check for unset parameter.
Let's say, we have a function setSalary that has one optional parameter. If the user does not supply the parameter then the default value of 10 should be used.
If you do the check like this:
function setSalary(dollars) {
salary = dollars || 10
}
This will give an unexpected result for a call like:
setSalary(0)
It will still set the 10 following the flow described above.
Double pipe operator
This example may be useful:
var section = document.getElementById('special');
if(!section){
section = document.getElementById('main');
}
It can also be:
var section = document.getElementById('special') || document.getElementById('main');
To add some explanation to all said before me, I should give you some examples to understand logical concepts.
var name = false || "Mohsen"; # name equals to Mohsen
var family = true || "Alizadeh" # family equals to true
It means if the left side evaluated as a true statement it will be finished and the left side will be returned and assigned to the variable. in other cases the right side will be returned and assigned.
And operator have the opposite structure like below.
var name = false && "Mohsen" # name equals to false
var family = true && "Alizadeh" # family equals to Alizadeh
Quote: "What does the construct x = x || y mean?"
Assigning a default value.
This means providing a default value of y to x,
in case x is still waiting for its value but hasn't received it yet or was deliberately omitted in order to fall back to a default.
And I have to add one more thing: This bit of shorthand is an abomination. It misuses an accidental interpreter optimization (not bothering with the second operation if the first is truthy) to control an assignment. That use has nothing to do with the purpose of the operator. I do not believe it should ever be used.
I prefer the ternary operator for initialization, for example,
var title = title?title:'Error';
This uses a one-line conditional operation for its correct purpose. It still plays unsightly games with truthiness but, that's JavaScript for you.
Optional Parameters
I often have JavaScript functions with optional parameters.
Instead of writing a long check like this:
if(param === null || param === undefined){
param = {};
}
I usually use the following syntax:
function doSomething(optionalParam, optionalCallback, optionalFlag){
optionalParam = optionalParam || {};
optionalParam["xyz"] = "value"; //Won't fail if the user didn't pass a value
optionalCallback = optionalCallback || function(){};
optionalCallback(); //If I need the function on 20 places, I don't have to insert a typeof-check everywhere
optionalFlag = optionalFlag || false;
}
The advantages are clear and I can deal with both undefined and null parameters.
This will, however, not work for optional flags that default to true:
someFlag = someFlag || true; //Will never evaluate to false.
Return values
Here's another example where I use that syntax:
function getValueOfIndex(idx){
return anArray[idx] || null; //Returns null if the index is out of bounds
}
My Question:
How does the || operator work in my use case?
Can I use || for all of these situations?
Is there any reason not to use it?
Are there any other types or values where this syntax will fail?
Edit: I recognised that my last point (difference between null||undefined and undefined||null) should belong to a seperate question, so I removed it.
The general answer is that you can't use
parameter = parameter || default;
if the user should be able to pass an explicit parameter that's falsey, and that should take precedence over the default. In that case you need to test explicitly for undefined:
parameter = typeof parameter == "undefined" ? default : parameter;
If the user should be able to pass an explicit undefined value and have that take precedence (a very perverse thing to do), you'll need to test arguments.length to determine how many arguments were passed, so you can default only the remainder.
Why do the following two expression return different values?
null || undefined //evaluates to undefined
undefined || null //evaluates to null
something || somethingElse is an expression, meaning it will always return a value, either the first truthy value, otherwise the last value at all. For example false || 17 is 17. Both null and undefined are falsy, so in both cases, || will return the last value.
Can I use || for all of these situations?
|| Can sometimes do something you wouldn't expect
function thisIsUnexpected (a) {
a = a || "other value";
reurn a;
}
thisIsUnexpected(); // will return "other value";
// it can fail when you pass in a falsy value
thisIsUnexpected(false); // you would want it to return false, yet it will return "other value"
To get the correct behavior, you'd want to use a ternary expression as follows. This can never fail.
function thisIsExpected (a) {
a = (a === undefined) ? "other value" : a;
reurn a;
}
The logical OR operator can be used as long as the expression used to test for the presence of an argument returns true for all the cases where an argument is supplied.
someFlag = someFlag || true; //Will never evaluate to false.
Your example above doesn't work because I can supply the argument false and the expression used to test for the presence of an argument (in this case simply someFlag) will still evaluate to false, implying that an argument was not specified.
So you would need to modify the code to something like:
someFlag = (typeof someFlag === 'boolean') ? someFlag : true;
This question already has answers here:
JavaScript check if variable exists (is defined/initialized)
(32 answers)
Closed 6 years ago.
I am trying to test to see whether a Javascript variable is undefined.
You will see that I am not expecting the value of predQuery[preId] to be 'undefined' if I don't first get an alert saying "its unbelievable". But I often do, so I am guessing that my statement
predQuery[preId]=='undefined')
is not matching the undefined elements properly.
if((predQuery.length < preId) || (predQuery[preId]=="") || (predQuery[preId]=='undefined')){
alert("its unbelievable");
alert(predQuery[preId]);
queryPreds[variables] = preId;
queryObjs[variables] = objId;
predQuery[preId] = variables;
}
else {
alert(predQuery[preId]);
var predIndex = predQuery[preId];
queryPreds[predIndex] = preId;
queryObjs[predIndex] = objId;
}
I can add more code if needed.
array[index] == 'undefined' compares the value of the array index to the string "undefined".
You're probably looking for typeof array[index] == 'undefined', which compares the type.
You are checking it the array index contains a string "undefined", you should either use the typeof operator:
typeof predQuery[preId] == 'undefined'
Or use the undefined global property:
predQuery[preId] === undefined
The first way is safer, because the undefined global property is writable, and it can be changed to any other value.
predQuery[preId]=='undefined'
You're testing against the string 'undefined'; you've confused this test with the typeof test which would return a string. You probably mean to be testing against the special value undefined:
predQuery[preId]===undefined
Note the strict-equality operator to avoid the generally-unwanted match null==undefined.
However there are two ways you can get an undefined value: either preId isn't a member of predQuery, or it is a member but has a value set to the special undefined value. Often, you only want to check whether it's present or not; in that case the in operator is more appropriate:
!(preId in predQuery)
There are more (many) ways to Rome:
//=>considering predQuery[preId] is undefined:
predQuery[preId] === undefined; //=> true
undefined === predQuery[preId] //=> true
predQuery[preId] || 'it\'s unbelievable!' //=> it's unbelievable
var isdef = predQuery[preId] ? predQuery[preId] : null //=> isdef = null
cheers!
Check for
if (predQuery[preId] === undefined)
Use the strict equal to operator. See comparison operators
try: typeof(predQuery[preId])=='undefined'
or more generally: typeof(yourArray[yourIndex])=='undefined'
You're comparing "undefined" to undefined, which returns false =)
This code works very well
function isUndefined(array, index) {
return ((String(array[index]) == "undefined") ? "Yes" : "No");
}