JavaScript strange behavior with operator precedence!!!1
-1
What do you think will be result of this expression?
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?NO!!! It's 15!
Now lets look at the other similar statement:
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Again no, now its 10. Why is that? I cant understand. But wait, there is two more:
var a = 10;
a = a + a++;
console.log(a);
// a = 21? Haha no! Its 20!And the last one:
var a = 10;
a = a++ + a;
console.log(a);
// 21))) now my mind is blown away (So can anybody tell, why javascript behaves this way?
javascript operators operator-precedence
edited Nov 14 '18 at 10:46
B001ᛦ
1,14251320
asked Nov 14 '18 at 10:20
Nikol LakinNikol Lakin
11
|
show 6 more comments
-1
What do you think will be result of this expression?
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?NO!!! It's 15!
Now lets look at the other similar statement:
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Again no, now its 10. Why is that? I cant understand. But wait, there is two more:
var a = 10;
a = a + a++;
console.log(a);
// a = 21? Haha no! Its 20!And the last one:
var a = 10;
a = a++ + a;
console.log(a);
// 21))) now my mind is blown away (So can anybody tell, why javascript behaves this way?
javascript operators operator-precedence
edited Nov 14 '18 at 10:46
B001ᛦ
1,14251320
asked Nov 14 '18 at 10:20
Nikol LakinNikol Lakin
11
I just wonder why they are correct, i dont say they aren't
– Nikol Lakin
Nov 14 '18 at 10:25
what about first two examples?
– Nikol Lakin
Nov 14 '18 at 10:27
1
While all the questions you have are grounded in the same basic issue (What order are operations carried out in when I assign a value to the same variable multiple times in the same expression? — which is a terrible idea in the first place ), I'm voting to close this as too broad because they are separate problems (and there are probably separate duplicates for each issue floating around anyway).
– Quentin
Nov 14 '18 at 10:28
1
More accessible references: stackoverflow.com/q/34094916/218196, stackoverflow.com/q/50869790/218196
– Felix Kling
Nov 14 '18 at 10:36
1
"The expresion in parentheses must evaluate first?" That's not what parenthesis mean. I guess one could say that they impact the order of evaluation (but that does not necessarily mean that they are evaluated first, just imagine you have multiple groups of parenthesis, they cannot be "first" all together). Rather they change the precedence of operands/operators and operators, which as a side effect impacts evaluation order. E.g.3 * 4 + 5is different than3 * (4 + 5). The LHS ofx * y, i.e.3is still evaluated first. The difference is in evaluating the RHS,4vs4 + 5.
– Felix Kling
Nov 14 '18 at 10:44
|
show 6 more comments
-1
-1
-1
What do you think will be result of this expression?
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?NO!!! It's 15!
Now lets look at the other similar statement:
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Again no, now its 10. Why is that? I cant understand. But wait, there is two more:
var a = 10;
a = a + a++;
console.log(a);
// a = 21? Haha no! Its 20!And the last one:
var a = 10;
a = a++ + a;
console.log(a);
// 21))) now my mind is blown away (So can anybody tell, why javascript behaves this way?
javascript operators operator-precedence
edited Nov 14 '18 at 10:46
B001ᛦ
1,14251320
asked Nov 14 '18 at 10:20
Nikol LakinNikol Lakin
11
What do you think will be result of this expression?
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?NO!!! It's 15!
Now lets look at the other similar statement:
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Again no, now its 10. Why is that? I cant understand. But wait, there is two more:
var a = 10;
a = a + a++;
console.log(a);
// a = 21? Haha no! Its 20!And the last one:
var a = 10;
a = a++ + a;
console.log(a);
// 21))) now my mind is blown away (So can anybody tell, why javascript behaves this way?
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? var a = 10;
a = a + a++;
console.log(a);
// a = 21? Haha no! Its 20!var a = 10;
a = a + a++;
console.log(a);
// a = 21? Haha no! Its 20!var a = 10;
a = a++ + a;
console.log(a);
// 21))) now my mind is blown away (var a = 10;
a = a++ + a;
console.log(a);
// 21))) now my mind is blown away (javascript operators operator-precedence
javascript operators operator-precedence
edited Nov 14 '18 at 10:46
B001ᛦ
1,14251320
asked Nov 14 '18 at 10:20
Nikol LakinNikol Lakin
11
edited Nov 14 '18 at 10:46
B001ᛦ
1,14251320
asked Nov 14 '18 at 10:20
Nikol LakinNikol Lakin
11
edited Nov 14 '18 at 10:46
B001ᛦ
1,14251320
edited Nov 14 '18 at 10:46
B001ᛦ
1,14251320
edited Nov 14 '18 at 10:46
B001ᛦ
1,14251320
1,14251320
asked Nov 14 '18 at 10:20
Nikol LakinNikol Lakin
11
asked Nov 14 '18 at 10:20
Nikol LakinNikol Lakin
11
asked Nov 14 '18 at 10:20
Nikol LakinNikol Lakin
11
11
I just wonder why they are correct, i dont say they aren't
– Nikol Lakin
Nov 14 '18 at 10:25
what about first two examples?
– Nikol Lakin
Nov 14 '18 at 10:27
1
While all the questions you have are grounded in the same basic issue (What order are operations carried out in when I assign a value to the same variable multiple times in the same expression? — which is a terrible idea in the first place ), I'm voting to close this as too broad because they are separate problems (and there are probably separate duplicates for each issue floating around anyway).
– Quentin
Nov 14 '18 at 10:28
1
More accessible references: stackoverflow.com/q/34094916/218196, stackoverflow.com/q/50869790/218196
– Felix Kling
Nov 14 '18 at 10:36
1
"The expresion in parentheses must evaluate first?" That's not what parenthesis mean. I guess one could say that they impact the order of evaluation (but that does not necessarily mean that they are evaluated first, just imagine you have multiple groups of parenthesis, they cannot be "first" all together). Rather they change the precedence of operands/operators and operators, which as a side effect impacts evaluation order. E.g.3 * 4 + 5is different than3 * (4 + 5). The LHS ofx * y, i.e.3is still evaluated first. The difference is in evaluating the RHS,4vs4 + 5.
– Felix Kling
Nov 14 '18 at 10:44
|
show 6 more comments
I just wonder why they are correct, i dont say they aren't
– Nikol Lakin
Nov 14 '18 at 10:25
what about first two examples?
– Nikol Lakin
Nov 14 '18 at 10:27
1
While all the questions you have are grounded in the same basic issue (What order are operations carried out in when I assign a value to the same variable multiple times in the same expression? — which is a terrible idea in the first place ), I'm voting to close this as too broad because they are separate problems (and there are probably separate duplicates for each issue floating around anyway).
– Quentin
Nov 14 '18 at 10:28
1
More accessible references: stackoverflow.com/q/34094916/218196, stackoverflow.com/q/50869790/218196
– Felix Kling
Nov 14 '18 at 10:36
1
"The expresion in parentheses must evaluate first?" That's not what parenthesis mean. I guess one could say that they impact the order of evaluation (but that does not necessarily mean that they are evaluated first, just imagine you have multiple groups of parenthesis, they cannot be "first" all together). Rather they change the precedence of operands/operators and operators, which as a side effect impacts evaluation order. E.g.3 * 4 + 5is different than3 * (4 + 5). The LHS ofx * y, i.e.3is still evaluated first. The difference is in evaluating the RHS,4vs4 + 5.
– Felix Kling
Nov 14 '18 at 10:44
I just wonder why they are correct, i dont say they aren't
– Nikol Lakin
Nov 14 '18 at 10:25
I just wonder why they are correct, i dont say they aren't
– Nikol Lakin
Nov 14 '18 at 10:25
what about first two examples?
– Nikol Lakin
Nov 14 '18 at 10:27
what about first two examples?
– Nikol Lakin
Nov 14 '18 at 10:27
1
1
While all the questions you have are grounded in the same basic issue (What order are operations carried out in when I assign a value to the same variable multiple times in the same expression? — which is a terrible idea in the first place ), I'm voting to close this as too broad because they are separate problems (and there are probably separate duplicates for each issue floating around anyway).
– Quentin
Nov 14 '18 at 10:28
While all the questions you have are grounded in the same basic issue (What order are operations carried out in when I assign a value to the same variable multiple times in the same expression? — which is a terrible idea in the first place ), I'm voting to close this as too broad because they are separate problems (and there are probably separate duplicates for each issue floating around anyway).
– Quentin
Nov 14 '18 at 10:28
1
1
More accessible references: stackoverflow.com/q/34094916/218196, stackoverflow.com/q/50869790/218196
– Felix Kling
Nov 14 '18 at 10:36
More accessible references: stackoverflow.com/q/34094916/218196, stackoverflow.com/q/50869790/218196
– Felix Kling
Nov 14 '18 at 10:36
1
1
"The expresion in parentheses must evaluate first?" That's not what parenthesis mean. I guess one could say that they impact the order of evaluation (but that does not necessarily mean that they are evaluated first, just imagine you have multiple groups of parenthesis, they cannot be "first" all together). Rather they change the precedence of operands/operators and operators, which as a side effect impacts evaluation order. E.g.
3 * 4 + 5 is different than 3 * (4 + 5). The LHS of x * y, i.e. 3 is still evaluated first. The difference is in evaluating the RHS, 4 vs 4 + 5.– Felix Kling
Nov 14 '18 at 10:44
"The expresion in parentheses must evaluate first?" That's not what parenthesis mean. I guess one could say that they impact the order of evaluation (but that does not necessarily mean that they are evaluated first, just imagine you have multiple groups of parenthesis, they cannot be "first" all together). Rather they change the precedence of operands/operators and operators, which as a side effect impacts evaluation order. E.g.
3 * 4 + 5 is different than 3 * (4 + 5). The LHS of x * y, i.e. 3 is still evaluated first. The difference is in evaluating the RHS, 4 vs 4 + 5.– Felix Kling
Nov 14 '18 at 10:44
|
show 6 more comments
2 Answers
2
active
oldest
votes
1
To understand this better, let us consider another variable
CASE 1
var a = 10;
a = a + (a = 5);
console.log(a); // 15
a = 10;
var b = a + (a = 5);
console.log(a); // 5
console.log(b); // 15a + (a = 5)
5 + (a = 5) // here a is assigned a value 5 and bracket returns 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 2
var a = 10;
a = (a = 5) + a;
console.log(a); // 10
a = 10;
var b = (a = 5) + a;
console.log(a); // 5
console.log(b); // 10(a = 5) + a
(a = 5) + a // here a is assigned a value 5 and bracket returns value 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 3
var a = 10;
a = a + a++;
console.log(a); // 20
a = 10;
var b = a + a++;
console.log(a); // 11
console.log(b); // 20a + a++ // post increment i.e. it will return the value first and then increment
10 + 10 // where value of a will 10 + 1 i.e. 11
20 // value will be assigned to variable on LHS
CASE 4
var a = 10;
a = a++ + a;
console.log(a); // 21
a = 10;
var b = a++ + a;
console.log(a); // 11
console.log(b); // 21a++ + a // post increment i.e. it will return the value first and then increment
10 + a // where value of a will 10 + 1 i.e. 11
10 + 11 // as value of a is 11
21 // value will be assigned to variable on LHS
answered Nov 14 '18 at 10:47
Nikhil AggarwalNikhil Aggarwal
24k32747
add a comment |
0
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?It's 15 because the value of a was 10 and you added (a = 5) after, which changed a's value to 5. Therefore, you did 10+5.
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Now that you have (a = 5) first, you're changing its value to 5 before adding a. Therefore, you're doing 5+5.
Hope this helps.
answered Nov 14 '18 at 10:47
Ryan WallsRyan Walls
6315
But the expression in parentheses must be evaluated first, i thought that what operator precedence mean
– Nikol Lakin
Nov 14 '18 at 11:16
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53297856%2fjavascript-strange-behavior-with-operator-precedence1%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
To understand this better, let us consider another variable
CASE 1
var a = 10;
a = a + (a = 5);
console.log(a); // 15
a = 10;
var b = a + (a = 5);
console.log(a); // 5
console.log(b); // 15a + (a = 5)
5 + (a = 5) // here a is assigned a value 5 and bracket returns 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 2
var a = 10;
a = (a = 5) + a;
console.log(a); // 10
a = 10;
var b = (a = 5) + a;
console.log(a); // 5
console.log(b); // 10(a = 5) + a
(a = 5) + a // here a is assigned a value 5 and bracket returns value 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 3
var a = 10;
a = a + a++;
console.log(a); // 20
a = 10;
var b = a + a++;
console.log(a); // 11
console.log(b); // 20a + a++ // post increment i.e. it will return the value first and then increment
10 + 10 // where value of a will 10 + 1 i.e. 11
20 // value will be assigned to variable on LHS
CASE 4
var a = 10;
a = a++ + a;
console.log(a); // 21
a = 10;
var b = a++ + a;
console.log(a); // 11
console.log(b); // 21a++ + a // post increment i.e. it will return the value first and then increment
10 + a // where value of a will 10 + 1 i.e. 11
10 + 11 // as value of a is 11
21 // value will be assigned to variable on LHS
answered Nov 14 '18 at 10:47
Nikhil AggarwalNikhil Aggarwal
24k32747
add a comment |
1
To understand this better, let us consider another variable
CASE 1
var a = 10;
a = a + (a = 5);
console.log(a); // 15
a = 10;
var b = a + (a = 5);
console.log(a); // 5
console.log(b); // 15a + (a = 5)
5 + (a = 5) // here a is assigned a value 5 and bracket returns 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 2
var a = 10;
a = (a = 5) + a;
console.log(a); // 10
a = 10;
var b = (a = 5) + a;
console.log(a); // 5
console.log(b); // 10(a = 5) + a
(a = 5) + a // here a is assigned a value 5 and bracket returns value 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 3
var a = 10;
a = a + a++;
console.log(a); // 20
a = 10;
var b = a + a++;
console.log(a); // 11
console.log(b); // 20a + a++ // post increment i.e. it will return the value first and then increment
10 + 10 // where value of a will 10 + 1 i.e. 11
20 // value will be assigned to variable on LHS
CASE 4
var a = 10;
a = a++ + a;
console.log(a); // 21
a = 10;
var b = a++ + a;
console.log(a); // 11
console.log(b); // 21a++ + a // post increment i.e. it will return the value first and then increment
10 + a // where value of a will 10 + 1 i.e. 11
10 + 11 // as value of a is 11
21 // value will be assigned to variable on LHS
answered Nov 14 '18 at 10:47
Nikhil AggarwalNikhil Aggarwal
24k32747
add a comment |
1
1
1
To understand this better, let us consider another variable
CASE 1
var a = 10;
a = a + (a = 5);
console.log(a); // 15
a = 10;
var b = a + (a = 5);
console.log(a); // 5
console.log(b); // 15a + (a = 5)
5 + (a = 5) // here a is assigned a value 5 and bracket returns 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 2
var a = 10;
a = (a = 5) + a;
console.log(a); // 10
a = 10;
var b = (a = 5) + a;
console.log(a); // 5
console.log(b); // 10(a = 5) + a
(a = 5) + a // here a is assigned a value 5 and bracket returns value 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 3
var a = 10;
a = a + a++;
console.log(a); // 20
a = 10;
var b = a + a++;
console.log(a); // 11
console.log(b); // 20a + a++ // post increment i.e. it will return the value first and then increment
10 + 10 // where value of a will 10 + 1 i.e. 11
20 // value will be assigned to variable on LHS
CASE 4
var a = 10;
a = a++ + a;
console.log(a); // 21
a = 10;
var b = a++ + a;
console.log(a); // 11
console.log(b); // 21a++ + a // post increment i.e. it will return the value first and then increment
10 + a // where value of a will 10 + 1 i.e. 11
10 + 11 // as value of a is 11
21 // value will be assigned to variable on LHS
answered Nov 14 '18 at 10:47
Nikhil AggarwalNikhil Aggarwal
24k32747
To understand this better, let us consider another variable
CASE 1
var a = 10;
a = a + (a = 5);
console.log(a); // 15
a = 10;
var b = a + (a = 5);
console.log(a); // 5
console.log(b); // 15a + (a = 5)
5 + (a = 5) // here a is assigned a value 5 and bracket returns 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 2
var a = 10;
a = (a = 5) + a;
console.log(a); // 10
a = 10;
var b = (a = 5) + a;
console.log(a); // 5
console.log(b); // 10(a = 5) + a
(a = 5) + a // here a is assigned a value 5 and bracket returns value 5
5 + 5 // here value of a is 5
10 // value will be assigned to variable on LHS
CASE 3
var a = 10;
a = a + a++;
console.log(a); // 20
a = 10;
var b = a + a++;
console.log(a); // 11
console.log(b); // 20a + a++ // post increment i.e. it will return the value first and then increment
10 + 10 // where value of a will 10 + 1 i.e. 11
20 // value will be assigned to variable on LHS
CASE 4
var a = 10;
a = a++ + a;
console.log(a); // 21
a = 10;
var b = a++ + a;
console.log(a); // 11
console.log(b); // 21a++ + a // post increment i.e. it will return the value first and then increment
10 + a // where value of a will 10 + 1 i.e. 11
10 + 11 // as value of a is 11
21 // value will be assigned to variable on LHS
var a = 10;
a = a + (a = 5);
console.log(a); // 15
a = 10;
var b = a + (a = 5);
console.log(a); // 5
console.log(b); // 15var a = 10;
a = a + (a = 5);
console.log(a); // 15
a = 10;
var b = a + (a = 5);
console.log(a); // 5
console.log(b); // 15var a = 10;
a = (a = 5) + a;
console.log(a); // 10
a = 10;
var b = (a = 5) + a;
console.log(a); // 5
console.log(b); // 10var a = 10;
a = (a = 5) + a;
console.log(a); // 10
a = 10;
var b = (a = 5) + a;
console.log(a); // 5
console.log(b); // 10var a = 10;
a = a + a++;
console.log(a); // 20
a = 10;
var b = a + a++;
console.log(a); // 11
console.log(b); // 20var a = 10;
a = a + a++;
console.log(a); // 20
a = 10;
var b = a + a++;
console.log(a); // 11
console.log(b); // 20var a = 10;
a = a++ + a;
console.log(a); // 21
a = 10;
var b = a++ + a;
console.log(a); // 11
console.log(b); // 21var a = 10;
a = a++ + a;
console.log(a); // 21
a = 10;
var b = a++ + a;
console.log(a); // 11
console.log(b); // 21answered Nov 14 '18 at 10:47
Nikhil AggarwalNikhil Aggarwal
24k32747
answered Nov 14 '18 at 10:47
Nikhil AggarwalNikhil Aggarwal
24k32747
answered Nov 14 '18 at 10:47
Nikhil AggarwalNikhil Aggarwal
24k32747
answered Nov 14 '18 at 10:47
Nikhil AggarwalNikhil Aggarwal
24k32747
24k32747
add a comment |
add a comment |
0
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?It's 15 because the value of a was 10 and you added (a = 5) after, which changed a's value to 5. Therefore, you did 10+5.
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Now that you have (a = 5) first, you're changing its value to 5 before adding a. Therefore, you're doing 5+5.
Hope this helps.
answered Nov 14 '18 at 10:47
Ryan WallsRyan Walls
6315
But the expression in parentheses must be evaluated first, i thought that what operator precedence mean
– Nikol Lakin
Nov 14 '18 at 11:16
add a comment |
0
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?It's 15 because the value of a was 10 and you added (a = 5) after, which changed a's value to 5. Therefore, you did 10+5.
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Now that you have (a = 5) first, you're changing its value to 5 before adding a. Therefore, you're doing 5+5.
Hope this helps.
answered Nov 14 '18 at 10:47
Ryan WallsRyan Walls
6315
But the expression in parentheses must be evaluated first, i thought that what operator precedence mean
– Nikol Lakin
Nov 14 '18 at 11:16
add a comment |
0
0
0
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?It's 15 because the value of a was 10 and you added (a = 5) after, which changed a's value to 5. Therefore, you did 10+5.
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Now that you have (a = 5) first, you're changing its value to 5 before adding a. Therefore, you're doing 5+5.
Hope this helps.
answered Nov 14 '18 at 10:47
Ryan WallsRyan Walls
6315
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?It's 15 because the value of a was 10 and you added (a = 5) after, which changed a's value to 5. Therefore, you did 10+5.
var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? Now that you have (a = 5) first, you're changing its value to 5 before adding a. Therefore, you're doing 5+5.
Hope this helps.
var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?var a = 10;
a = a + (a = 5);
console.log(a);
// a = 10?var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? var a = 10;
a = (a = 5) + a;
console.log(a);
// a = 15? answered Nov 14 '18 at 10:47
Ryan WallsRyan Walls
6315
answered Nov 14 '18 at 10:47
Ryan WallsRyan Walls
6315
answered Nov 14 '18 at 10:47
Ryan WallsRyan Walls
6315
answered Nov 14 '18 at 10:47
Ryan WallsRyan Walls
6315
6315
But the expression in parentheses must be evaluated first, i thought that what operator precedence mean
– Nikol Lakin
Nov 14 '18 at 11:16
add a comment |
But the expression in parentheses must be evaluated first, i thought that what operator precedence mean
– Nikol Lakin
Nov 14 '18 at 11:16
But the expression in parentheses must be evaluated first, i thought that what operator precedence mean
– Nikol Lakin
Nov 14 '18 at 11:16
But the expression in parentheses must be evaluated first, i thought that what operator precedence mean
– Nikol Lakin
Nov 14 '18 at 11:16
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53297856%2fjavascript-strange-behavior-with-operator-precedence1%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I just wonder why they are correct, i dont say they aren't
– Nikol Lakin
Nov 14 '18 at 10:25
what about first two examples?
– Nikol Lakin
Nov 14 '18 at 10:27
1
While all the questions you have are grounded in the same basic issue (What order are operations carried out in when I assign a value to the same variable multiple times in the same expression? — which is a terrible idea in the first place ), I'm voting to close this as too broad because they are separate problems (and there are probably separate duplicates for each issue floating around anyway).
– Quentin
Nov 14 '18 at 10:28
1
More accessible references: stackoverflow.com/q/34094916/218196, stackoverflow.com/q/50869790/218196
– Felix Kling
Nov 14 '18 at 10:36
1
"The expresion in parentheses must evaluate first?" That's not what parenthesis mean. I guess one could say that they impact the order of evaluation (but that does not necessarily mean that they are evaluated first, just imagine you have multiple groups of parenthesis, they cannot be "first" all together). Rather they change the precedence of operands/operators and operators, which as a side effect impacts evaluation order. E.g.
3 * 4 + 5is different than3 * (4 + 5). The LHS ofx * y, i.e.3is still evaluated first. The difference is in evaluating the RHS,4vs4 + 5.– Felix Kling
Nov 14 '18 at 10:44