how can I turn two statements into a ternary - javascript

I have two if statements that'd like to turn into a ternary but I don't know how.
if (val === true && optval === 'car' || val === true && optval === 'truck' )view_list.style.display = 'none';
if (val === true && optval === 'buss' || val === true && optval === 'van')view_list.style.display = '';

Apart in hacks I won't mention, a ternary is used when you want to return a value in all cases.
If val isn't true, then you do nothing.
Therefore this can't be used here.

view_list.style.display =
(val === true && optval === 'car' ||
val === true && optval === 'truck' ) ?
'none' :
((val === true && optval === 'buss' ||
val === true && optval === 'van') ?
'' :
view_list.style.display ) ;
or more succinctly :
view_list.style.display =
(val === true && ( optval === 'car' || optval === 'truck' ) ) ?
'none' :
((val === true && ( optval === 'buss' || optval === 'van' ) ?
'' :
view_list.style.display );
The parentheses around the second conditional are optional.
Of course this is not really a good approach, and this answer should not be seen as an endorsement to this style of programming.

Because nobody has said it yet, switch
if (val === true)
switch (optval) {
case 'car':
case 'truck':
view_list.style.display = 'none';
break;
case 'buss':
case 'van':
view_list.style.display = '';
break;
}

Not what you were asking for, but as #dystroy said, this is not the best case for a ternary.
But you can still make it a bit more elegant. How about
var styleMapping = {
'car': 'none',
'truck': 'none',
'buss': '',
'van': ''
};
// Apparently val is always required in your example
if (val) {
view_list.style.display = styleMapping[optval];
}
This also has the benefit that refactoring and adding new styles is very easy without having to write much code. Just add one line in the mapping and you're done. And of course it's much more readable and closer to what you intend to do.
You don't really care about boolean logic, you just want to have a different style based on what optval is set to.

Related

if statement with javascript condition operator

just have 2 question regarding JS conditional operator, is the below 2 expression valid?
1.
if(isUser && isUser === true || isGuest && isGuest === true){
//...
}
I am wondering do I have to add () to make it like and still have the same functioning:
if((isUser && isUser === true) || (isGuest && isGuest === true)){
//...
}
const items = list.orderList && list.orderList.isUser === true || list.orderList.isGuest ? list.items : [];
I am wondering do I have to add () to make it like and functioning the same as above conditional operator:
const items = list.orderList && (list.orderList.isUser === true || list.orderList.isGuest === true) ? list.items : [];
As per Operator Precedence in the MDN docs, logical AND takes precedence over logical OR. Therefore,
expression1 || expression2 && expression3
will evaluate to
expression1 || (expression2 && expression3)
Therefore,
isUser && isUser === true || isGuest && isGuest === true
naturally evaluates to
(isUser && isUser === true) || (isGuest && isGuest === true)
anyway, so you do not need parentheses..
But since, in your second example, you want to evaluate OR then AND, you do need parentheses for it to evaluate the way you require, as
list.orderList && list.orderList.isUser === true || list.orderList.isGuest
will evaluate to
(list.orderList && list.orderList.isUser === true) || list.orderList.isGuest

How can I test two strings for equivalence in JavaScript, considering null and empty string the same?

If I compare "a" and "b", that should be false.
If I compare "a" and "a", that should be true.
If I compare "" and null, that should be true.
I could write my own method, but thought there was perhaps a JavaScript shortcut.
Edit: I was thinking something like this:
areDbSame(s1, s2) {
if (s1 === null) s1 = "";
if (s2 === null) s2 = "";
return s1 === s2;
}
Edit2: Settled on this version:
areDbSame(s1, s2) {
return (s1 === null ? "" : s1) === (s2 === null ? "" : s2);
}
Just before you test the equality of your string, you could do a simple one line enforcement, by converting to '' in the case of null. For example (if you also don't care about undefined, false, etc):
// testString becomes the one you are testing
var testString = myString || '';
If you only want to ensure null is blank
var testString = (myString === null) ? '' : myString;
Then you can simply do your string comparisons using testString, and not worry about the null equalities.
IMO this is the cleanest answer because it doesn't convolute the original equality testing of javascript strings. It is the same as saying, let's split the problem up into two parts.
1) When should my string be considered blank, and
2) Now I can just check for regular string equality.
function areEqualStrings(a, b) {
var otherEqualValues = ['', null];
if(typeof a === 'string' && typeof b === 'string') {
return a === b;
} else if(otherEqualValues.indexOf(a) > -1 && otherEqualValues.indexOf(b) > -1) {
return !a === !b;
} else {
return false;
}
}
When coercing JavaScript values, !null is true and !'' is true, so those would result in being equal.
Here's the test (screenshotted from my console):
This function should do it. It type checks first and short circuits otherwise.
function stringCompare(a, b) {
if (((a === null || typeof a === 'string') ||
(b === null || typeof b === 'string')) &&
((a === '' && b === null) ||
(b === '' && a === null) ||
(a === b))) {
return true;
}
return false;
}
No it hasn`t. The two first cases you can do naturally using operator =.
The third case it is impossible because "" is considered a empty string and null has any type. So they never can be true naturally. To do this, you have to write your own method.
Just to be clear. You can use operators = (equal) to do comparison:
== equal to
`x == 8 false
x == 5 true
x == "5" true
=== equal value and equal type
x === 5 true
x === "5" false
Hope it helps

Alternative to nested ternary operator in JS [closed]

Closed. This question is opinion-based. It is not currently accepting answers.
Want to improve this question? Update the question so it can be answered with facts and citations by editing this post.
Closed 8 months ago.
Improve this question
I personally love ternary operators, and in my humble opinion, they make complicated expressions very easy to digest. Take this one:
const word = (distance === 0) ? 'a'
: (distance === 1 && diff > 3) ? 'b'
: (distance === 2 && diff > 5 && key.length > 5) ? 'c'
: 'd';
However in our project's ESLINT rules nested ternary operators are forbidden, so I have to get rid of the above.
I'm trying to find out alternatives to this approach. I really don't want to turn it into a huge if / else statement, but don't know if there's any other options.
Your alternatives here are basically:
That if/else you don't want to do
A switch combined with if/else
I tried to come up with a reasonable lookup map option, but it got unreasonable fairly quickly.
I'd go for #1, it's not that big:
if (res.distance == 0) {
word = 'a';
} else if (res.distance == 1 && res.difference > 3) {
word = 'b';
} else if (res.distance == 2 && res.difference > 5 && String(res.key).length > 5) {
word = 'c';
} else {
word = 'd';
}
If all the braces and vertical size bother you, without them it's almost as concise as the conditional operator version:
if (res.distance == 0) word = 'a';
else if (res.distance == 1 && res.difference > 3) word = 'b';
else if (res.distance == 2 && res.difference > 5 && String(res.key).length > 5) word = 'c';
else word = 'd';
(I'm not advocating that, I never advocate leaving off braces or putting the statement following an if on the same line, but others have different style perspectives.)
#2 is, to my mind, more clunky but that's probably more a style comment than anything else:
word = 'd';
switch (res.distance) {
case 0:
word = 'a';
break;
case 1:
if (res.difference > 3) {
word = 'b';
}
break;
case 2:
if (res.difference > 5 && String(res.key).length > 5) {
word = 'c';
}
break;
}
And finally, and I am not advocating this, you can take advantage of the fact that JavaScript's switch is unusual in the B-syntax language family: The case statements can be expressions, and are matched against the switch value in source code order:
switch (true) {
case res.distance == 0:
word = 'a';
break;
case res.distance == 1 && res.difference > 3:
word = 'b';
break;
case res.distance == 2 && res.difference > 5 && String(res.key).length > 5:
word = 'c';
break;
default:
word = 'd';
break;
}
How ugly is that? :-)
To my taste, a carefully structured nested ternary beats all those messy ifs and switches:
const isFoo = res.distance === 0;
const isBar = res.distance === 1 && res.difference > 3;
const isBaz = res.distance === 2 && res.difference > 5 && String(res.key).length > 5;
const word =
isFoo ? 'a' :
isBar ? 'b' :
isBaz ? 'c' :
'd' ;
You could write an immediately invoked function expression to make it a little more readable:
const word = (() => {
if (res.distance === 0) return 'a';
if (res.distance === 1 && res.difference > 3) return 'b';
if (res.distance === 2 && res.difference > 5 && String(res.key).length > 5) return 'c';
return 'd';
})();
Link to repl
We can simplify it using basic operators like && and ||
let obj = {}
function checkWord (res) {
return (res.distance === 0) && 'a'
|| (res.distance === 1 && res.difference > 3) && 'b'
|| (res.distance === 2 && res.difference > 5 && String(res.key).length > 5) && 'c'
|| 'd';
}
// case 1 pass
obj.distance = 0
console.log(checkWord(obj))
// case 2 pass
obj.distance = 1
obj.difference = 4
console.log(checkWord(obj))
// case 3 pass
obj.distance = 2
obj.difference = 6
obj.key = [1,2,3,4,5,6]
console.log(checkWord(obj))
// case 4 fail all cases
obj.distance = -1
console.log(checkWord(obj))
If you are looking to use const with a nested ternary expression, you can replace the ternary with a function expression.
const res = { distance: 1, difference: 5 };
const branch = (condition, ifTrue, ifFalse) => condition?ifTrue:ifFalse;
const word = branch(
res.distance === 0, // if
'a', // then
branch( // else
res.distance === 1 && res.difference > 3, // if
'b', // then
branch( // else
res.distance === 2 && res.difference > 5, // if
'c', // then
'd' // else
)
)
);
console.log(word);
or using named parameters via destructuring...
const branch2 = function(branch) {
return branch.if ? branch.then : branch.else;
}
const fizzbuzz = function(num) {
return branch2({
if: num % 3 === 0 && num % 5 === 0,
then: 'fizzbuzz',
else: branch2({
if: num % 3 === 0,
then: 'fizz',
else: branch2({
if: num % 5 === 0,
then: 'buzz',
else: num
})
})
});
}
console.log(
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16].map(
cv => fizzbuzz(cv)
)
);
edit
It may be clearer to model it after the python if expression like this:
const res = { distance: 1, difference: 5 };
const maybe = def => ({
if: expr => {
if (expr) {
return { else: () => def };
} else {
return { else: els => els };
}
}
});
const word = maybe('a').if(res.distance === 0).else(
maybe('b').if(res.distance === 1 && res.difference > 3).else(
maybe('c').if(res.distance === 2 && res.difference > 5).else('d')
)
);
console.log(word);
edit
Another edit to remove the nested if/else branches:
const res = { distance: 1, difference: 5 };
const makeResolvedValue = def => {
const elseProp = () => def;
return function value() {
return {
if: () => ({ else: elseProp, value })
};
}
};
const value = def => ({
if: expr => {
if (expr) {
return { else: () => def, value: makeResolvedValue(def) };
} else {
return { else: els => els, value };
}
}
});
// with branching if needed
const word = value('a').if(res.distance === 0)
.else(value('b').if(res.distance === 1 && res.difference > 3)
.else(value('c').if(res.distance === 2 && res.difference > 5)
.else('d')
)
);
console.log(word)
// implicit else option for clarity
const word2 = value('a').if(res.distance === 0)
.value('b').if(res.distance === 1 && res.difference > 3)
.value('c').if(res.distance === 2 && res.difference > 5)
.else('d');
console.log(word2);
If all your truthy conditions evaluate to truthy values (so the value between the question mark and the semicolon evaluates to true if coerced to boolean...) you could make your ternary expressions return false as the falsy expression. Then you could chain them with the bitwise or (||) operator to test the next condition, until the last one where you return the default value.
In the example below, the "condsXXX" array represent the result of evaluating the conditions. "conds3rd" simulates the 3rd condition is true and "condsNone" simulates no condition is true. In a real life code, you'd have the conditions "inlined" in the assignment expression:
var conds3rd = [false, false, true];
var condsNone = [false, false, false];
var val3rd = (conds3rd[0] ? 1 : false) ||
(conds3rd[1] ? 2 : false) ||
(conds3rd[2] ? 3 : 4);
var valNone = (condsNone[0] ? 1 : false) ||
(condsNone[1] ? 2 : false) ||
(condsNone[2] ? 3 : 4);
alert(val3rd);
alert(valNone);
Your example could end up like below:
word = ((res.distance === 0) ? 'a' : false) ||
((res.distance === 1 && res.difference > 3) ? 'b' : false) ||
((res.distance === 2 && res.difference > 5 && String(res.key).length > 5) ? 'c' : 'd';
As a side note, I don't feel it's a good looking code, but it is quite close to using the pure ternary operator like you aspire to do...
word = (res.distance === 0) ? 'a'
: (res.distance === 1 && res.difference > 3) ? 'b'
: (res.distance === 2 && res.difference > 5 && String(res.key).length > 5) ? 'c'
: 'd';
This is an older question, but this is how I would do it... I would start with the default case and then change the variable or pass it unchanged as desired.
var word = 'd';
word = (res.distance === 0) ? 'a' : word;
word = (res.distance === 1 && res.difference > 3) ? 'b' : word
word = (res.distance === 2 && res.difference > 5 && String(res.key).length > 5) ? 'c' : word;
Sometimes we have (or just love) to use one-line expressions or variable definitions.
So, we can use a combination of destructive assignments with the ternary operator. For example,
was:
const a = props.a ? props.a : cond2 ? 'val2.0' : 'val2.1' ;
let's update to:
const { a = cond2 ? 'val2.0' : 'val2.1' } = props;
It even remains relatively well readable.
I personally love using ternary expressions for one liners.
Although, I have to agree that nesting ternary expressions can lead to sketchy code.
I started playing with the Object constructor recently to write clearer code:
let param: "one" | "two" | "three";
// Before
let before: number = param === "one" ? 1 : param === "two" ? 2 : 3;
// After
let after: number = Object({
one: 1,
two: 2,
three: 3
})[param];
Real life example:
const opacity =
Platform.OS === "android"
? 1
: Object({
disabled: 0.3,
pressed: 0.7,
default: 1,
})[(disabled && "disabled") || (pressed && "pressed") || "default"];
If you use lodash you can use _.cond
Point free version with lodash/fp:
const getWord = _.cond([
[_.flow(_.get('distance'), _.eq(0)), _.constant('a')],
[_.flow(_.get('distance'), _.eq(1)) && _.flow(_.get('difference'), _.gt(3)), _.constant('b')],
[
_.flow(_.get('distance'), _.eq(2))
&& _.flow(_.get('difference'), _.gt(5))
&& _.flow(_.get('key'), _.toString, _.gt(5)),
_.constant('c'),
],
[_.stubTrue, _.constant('d')],
]);
If you're in the mood for something a little less readable... this might be for you. Write a general function to take an array of conditions (in the order you'd write your if/else) and an array of assignment values. Use .indexOf() to find the first truth in your conditions, and return the assignment array value at that index. Order is critical, conditions need to match up by index to the assignment you want:
const conditionalAssignment = (conditions, assignmentValues) => assignmentValues[conditions.indexOf(true)];
You can modify to handle truthy instead of struct true, and beware the undefined return if indexOf is -1
I faced this too recently and a google search led me here, and I want to share something I discovered recently regarding this:
a && b || c
is almost the same thing as
a ? b : c
as long as b is truthy. If b isn't truthy, you can work around it by using
!a && c || b
if c is truthy.
The first expression is evaluated as (a && b) || c as && has more priority than ||.
If a is truthy then a && b would evaluate to b if b is truthy, so the expression becomes b || c which evaluates to b if it is truthy, just like a ? b : c would if a is truthy, and if a is not truthy then the expression would evaluate to c as required.
Alternating between the && and || trick and ? and || in the layers of the statement tricks the no-nested-ternary eslint rule, which is pretty neat (although I would not recommend doing so unless there is no other way out).
A quick demonstration:
true ? false ? true : true ? false : true ? true ? true : false : true : true
// which is interpreted as
true ? (false ? true : (true ? false : (true ? (true ? true : false) : true))) : true
// now with the trick in alternate levels
true ? (false && true || (true ? false : (true && (true ? true : false) || true))) : true
// all of these evaluate to false btw
I actually cheated a bit by choosing an example where b is always truthy, but if you are just setting strings then this should work fine as even '0' is ironically truthy.
I've been using a switch(true) statement for these cases. In my opinion this syntax feels slightly more elegant than nested if/else operators
switch (true) {
case condition === true :
//do it
break;
case otherCondition === true && soOn < 100 :
// do that
break;
}
ES6 opens the door to this, a different take on a switch statement.
Object.entries({
['a']: res.distance === 0,
['b']: res.distance === 1 && res.difference > 3,
['c']: (res.distance === 2 && res.difference > 5 && String(res.key).length > 5) ? 'c'
}).filter(n => n[1] === true)[0][0]
I prefer inlined if-else in ternary expression
This..
let assignment;
if (loggedUser === onboard.Employee)
assignment = AssignmentEnum.Employee;
else if (loggedUser === onboard.CreatedBy)
assignment = AssignmentEnum.Manager;
else
assignment = 0;
..can be shortened to:
const assignment =
loggedUser === onboard.Employee ?
AssignmentEnum.Employee
:loggedUser === onboard.CreatedBy ?
AssignmentEnum.Manager
:
0;
Just add the following ES6 lint rules to allow the above code structure
On "indent", add this rule:
"ignoredNodes": ["ConditionalExpression"]
On "operator-linebreak", add this rule:
{ "overrides": { "?": "after", ":": "ignore" } }
But if you don't want to put the assigned value in their own line, aside from the settings above, add the multiline-ternary to off
"multiline-ternary": "off"
Here's an option that probably accomplishes the same thing, but might be verging on unreadable and liable to break. It's probably slightly less efficient than the ternary, too.
const word = [
{ value: 'a', when: () => distance === 0 },
{ value: 'b', when: () => distance === 1 && diff > 3 },
{ value: 'c', when: () => distance === 2 && diff > 5 && key.length > 5 },
{ value: 'd', when: () => 'otherwise' }, // just has to return something truthy
].find(({ when }) => when()).value
Array.prototype.find evaluates each element in the array until its callback returns a truthy value. Then it stops and returns, so we wouldn't evaluate any more cases than necessary.
We could also do it without a bunch of tiny functions. Slightly easier to read, but potentially less efficient if your calculations are slow or prone to throwing errors.
const word = [
{ value: 'a', when: distance === 0 },
{ value: 'b', when: distance === 1 && diff > 3 },
{ value: 'c', when: distance === 2 && diff > 5 && key.length > 5 },
{ value: 'd', when: 'otherwise' }, // just has to be something truthy
].find(({ when }) => when).value
But honestly, for shorter ternaries, I'd probably just go with an if/else (or remove the lint rule if you can get the rest of your team to agree).
I think some of the answers here kinda missed the point why the OP want to use ternary operator. Many prefers ternary expression as it allows them to DRY the assignment operations, not just because they can inline the returned value on same line as the condition and make the code have a lookup-like operation
If you deemed the switch(true) { is not ugly, I wish passing true is optional in javascript, either switch() { or switch {, then you can certainly DRY your assignment operations with switch:
const word =
(() => {
switch (true) {
case res.distance == 0:
return 'a';
case res.distance == 1 && res.difference > 3:
return 'b';
case res.distance == 2 && res.difference > 5 && String(res.key).length > 5:
return 'c';
default:
return 'd';
}
})();
Beside if-else and switch, you can also try square brackets.
const testName = cond1 ? [cond2 : a : b] : [cond3 ? c : d]

Which Logic Operator Takes Precedence

So, I'm looking into writing a slightly more complex operation with logic operators in an if-else statement. I know I can do parentheses, and I know it's the better way of doing this, but I've gotten curious and so I'm going to ask. If I were to do something like this:
if (firstRun == true || selectedCategory != undefined && selectedState != undefined) {
//Do something
} else {
//Do something else
}
How will that be operated without the use of parentheses? I know there is an order of operations for logic operators, similar to PEMDAS, right? I'm curious if it'll be ran something like this:
firstRun == true || (selectedCategory != undefined && selectedState != undefined)
or maybe if the 'OR' operator takes precedence instead and it ends up going like:
(firstRun == true || selectedCategory != undefined) && selectedState != undefined
The full list would be nice, if you can find it somewhere, of the order of operations for this. Thanks!
My rule of thumb, which covers basically 99% of all use cases for conditional statements, is:
Grouping: ()
Member access . or [...]
Not: !
Comparison, e.g. < , >= , === , !=, ...
Logical AND &&
Logical OR ||
MDN gives you the exhaustive breakdown: JavaScript Operator Precedence
So for your example:
(firstRun == true || selectedCategory != undefined && selectedState != undefined)
equals
(firstRun == true) || ((selectedCategory != undefined) && (selectedState != undefined))
For anything more complex than the above mentioned cases, I would look into refactoring the code for readability's sake anyway!
There is a pretty good rule of thumb to this. Think of these operators as of mathematical ones:
AND is multiplication (eg. 0 * 1 = 0 => FALSE)
OR is adding (eg. 0 + 1 = 1 => TRUE)
When you remember this, all you have to know is that multiplication always comes before addition.
See this chart for precedence.
I'm not going to explain what happens because the next guy reading your code will think: "WTF? Does that do what it should?"
So the better solution is to wrap the terms in parentheses even if you know the precedence, applied it correctly and the code works
This follows the old wisdom that you shouldn't do everything you can just because you can do it. Always keep an eye on the consequences.
See Operator precedence.
&& is before ||, so your expression is equivalent to:
firstRun == true || (selectedCategory != undefined && selectedState != undefined)
It will be the first:
firstRun == true || (selectedCategory != undefined && selectedState != undefined)
As a general rule, in most programming languages, AND has higher precedence.
While logical operator precedence is not actually defined in the ECMAScript Specification, MDN does a pretty good job of it and even has a separate page for logical operators.
My concern I suppose, since logical operator precedence is not actually defined in the ECMAScript specification, each individual browser vendor can potentially be different (I'm talking to you, Internet Explorer!), so your mileage may vary.
In the event anyone wants to test this across different browsers, here's a test case fiddle: http://jsfiddle.net/HdzXq/
$(document).ready(function() {
function log(test) {
$('div#out').append($('<p />').text(test));
}
function testOperatorPrecedence() {
log('(false || false && false) === ' + (false || false && false).toString());
log('(false || false && true) === ' + (false || false && true).toString());
log('(false || true && false) === ' + (false || true && false).toString());
log('(false || true && true) === ' + (false || true && true).toString());
log('(true || false && false) === ' + (true || false && false).toString());
log('(true || false && true) === ' + (true || false && true).toString());
log('(true || true && false) === ' + (true || true && false).toString());
log('(true || true && true) === ' + (true || true && true).toString());
log('----------------------------');
log('(false || (false && false)) === ' + (false || (false && false)).toString());
log('(false || (false && true )) === ' + (false || (false && true)).toString());
log('(false || (true && false)) === ' + (false || (true && false)).toString());
log('(false || (true && true )) === ' + (false || (true && true)).toString());
log('(true || (false && false)) === ' + (true || (false && false)).toString());
log('(true || (false && true )) === ' + (true || (false && true)).toString());
log('(true || (true && false)) === ' + (true || (true && false)).toString());
log('(true || (true && true )) === ' + (true || (true && true)).toString());
}
testOperatorPrecedence();
});

What is the correct syntax for this 'OR' and 'AND' in this 'IF' statement?

I've got an 'if' statement and just wanted to know if these are both valid (which I believe they are) and if so what is the difference?
var type;
var type2;
if ((type == 'BOS'|| type == 'BPS'|| type == 'BRS') && (type2 == 'BOS'|| type2 == 'BPS'|| type2 == 'BRS))
OR
if ((type == 'BOS') || (type == 'BPS') || (type == 'BRS') && (type2 == 'BOS') || (type2 == 'BPS') || (type2 == 'BRS'))
Which has the correct syntax and do they do anything differently? is there a way to shorten this statement?
Thanks
The two statements are different. Yes, they are both valid statements, syntactically, but logically they differ. Since the && operator has a higher precedence than the || in javscript,
the resulting logic will evaluate as follows in statement 2:
1) (type == 'BRS') && (type2 == 'BOS')
2) (type == 'BOS') || (type == 'BPS') || (result of 1) || (type2 == 'BPS') || (type2 == 'BRS')
While in statement 1:
1) (type == 'BOS'|| type == 'BPS'|| type == 'BRS')
2) (type2 == 'BOS'|| type2 == 'BPS'|| type2 == 'BRS')
3) (result of 1) && (result of 2)
var type1;
var type2;
var correct_value = {
BRS: 1,
BOS: 1,
BPS: 1
};
if( correct_value[type1] && correct_value[type2]) {
alert('ok');
}
else {
alert('not ok');
}
Both conditional statement are valid but the result may be different.
The first statement will evaluate the OR value in the first bracket, and then evaluate the OR value in the second bracket, and finally evaluate the AND operator.
The second statement will evaluate the AND first and then evaluate from left to right without specific precedence (as in the first one).
You have to know which one do you actually use to determine the best refactoring code for it.
See this link: Operator Precedence.

Categories