Why is x == (x = y) not the same as (x = y) == x?
up vote
14
down vote
favorite
Consider the following example:
class Quirky {
public static void main(String args) {
int x = 1;
int y = 3;
System.out.println(x == (x = y)); // false
x = 1; // reset
System.out.println((x = y) == x); // true
}
}
I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.
Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.
java variable-assignment operator-precedence jls
edited 1 hour ago
Boann
36.6k1287121
asked 9 hours ago
John McClane
447114
|
show 2 more comments
up vote
14
down vote
favorite
Consider the following example:
class Quirky {
public static void main(String args) {
int x = 1;
int y = 3;
System.out.println(x == (x = y)); // false
x = 1; // reset
System.out.println((x = y) == x); // true
}
}
I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.
Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.
java variable-assignment operator-precedence jls
edited 1 hour ago
Boann
36.6k1287121
asked 9 hours ago
John McClane
447114
11
The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
9 hours ago
@LouisWasserman The term evaluate is, in my opinion, inapplicable here becausexdoes not need to be evaluated, it just loaded from memory.
– John McClane
9 hours ago
3
Evaluating the expressionx = yis certainly relevant, and causes the side effect thatxis set to the value ofy.
– Louis Wasserman
9 hours ago
@LouisWasserman Sure. I knew that, and it was the main idea that lead me toquickReplaceAndCompare().
– John McClane
9 hours ago
7
Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
6 hours ago
|
show 2 more comments
up vote
14
down vote
favorite
up vote
14
down vote
favorite
Consider the following example:
class Quirky {
public static void main(String args) {
int x = 1;
int y = 3;
System.out.println(x == (x = y)); // false
x = 1; // reset
System.out.println((x = y) == x); // true
}
}
I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.
Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.
java variable-assignment operator-precedence jls
edited 1 hour ago
Boann
36.6k1287121
asked 9 hours ago
John McClane
447114
Consider the following example:
class Quirky {
public static void main(String args) {
int x = 1;
int y = 3;
System.out.println(x == (x = y)); // false
x = 1; // reset
System.out.println((x = y) == x); // true
}
}
I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.
Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.
java variable-assignment operator-precedence jls
java variable-assignment operator-precedence jls
edited 1 hour ago
Boann
36.6k1287121
asked 9 hours ago
John McClane
447114
edited 1 hour ago
Boann
36.6k1287121
asked 9 hours ago
John McClane
447114
edited 1 hour ago
Boann
36.6k1287121
edited 1 hour ago
Boann
36.6k1287121
edited 1 hour ago
Boann
36.6k1287121
36.6k1287121
asked 9 hours ago
John McClane
447114
asked 9 hours ago
John McClane
447114
asked 9 hours ago
John McClane
447114
447114
11
The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
9 hours ago
@LouisWasserman The term evaluate is, in my opinion, inapplicable here becausexdoes not need to be evaluated, it just loaded from memory.
– John McClane
9 hours ago
3
Evaluating the expressionx = yis certainly relevant, and causes the side effect thatxis set to the value ofy.
– Louis Wasserman
9 hours ago
@LouisWasserman Sure. I knew that, and it was the main idea that lead me toquickReplaceAndCompare().
– John McClane
9 hours ago
7
Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
6 hours ago
|
show 2 more comments
11
The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
9 hours ago
@LouisWasserman The term evaluate is, in my opinion, inapplicable here becausexdoes not need to be evaluated, it just loaded from memory.
– John McClane
9 hours ago
3
Evaluating the expressionx = yis certainly relevant, and causes the side effect thatxis set to the value ofy.
– Louis Wasserman
9 hours ago
@LouisWasserman Sure. I knew that, and it was the main idea that lead me toquickReplaceAndCompare().
– John McClane
9 hours ago
7
Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
6 hours ago
11
11
The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
9 hours ago
The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
9 hours ago
@LouisWasserman The term evaluate is, in my opinion, inapplicable here because
x does not need to be evaluated, it just loaded from memory.– John McClane
9 hours ago
@LouisWasserman The term evaluate is, in my opinion, inapplicable here because
x does not need to be evaluated, it just loaded from memory.– John McClane
9 hours ago
3
3
Evaluating the expression
x = y is certainly relevant, and causes the side effect that x is set to the value of y.– Louis Wasserman
9 hours ago
Evaluating the expression
x = y is certainly relevant, and causes the side effect that x is set to the value of y.– Louis Wasserman
9 hours ago
@LouisWasserman Sure. I knew that, and it was the main idea that lead me to
quickReplaceAndCompare().– John McClane
9 hours ago
@LouisWasserman Sure. I knew that, and it was the main idea that lead me to
quickReplaceAndCompare().– John McClane
9 hours ago
7
7
Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
6 hours ago
Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
6 hours ago
|
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
25
down vote
== is a binary equality operator.
The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First
answered 8 hours ago
Andrew Tobilko
24.4k84182
1
@JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for==. Without the parentheses, neither of these expressions would be syntactically correct.
– Andrew Tobilko
7 hours ago
add a comment |
up vote
15
down vote
As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.
//the example values
x = 1;
y = 2;
So in quickReplaceAndCompare the following is done:
x == (x = y)
1 == (x = y)
1 == (x = 2) //assign 2 to x, returns 2
1 == 2
false
and in badReplaceAndCompare():
(x = y) == x
(x = 2) == x //assign 2 to x, returns 2
2 == x
2 == 2
true
note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.
answered 8 hours ago
Poohl
33319
add a comment |
up vote
0
down vote
In the first test you're checking does 1 == 3.
In the second test your checking does 3 == 3.
(x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?
In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?
answered 39 mins ago
Michael Puckett II
3,67541436
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
25
down vote
== is a binary equality operator.
The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First
answered 8 hours ago
Andrew Tobilko
24.4k84182
1
@JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for==. Without the parentheses, neither of these expressions would be syntactically correct.
– Andrew Tobilko
7 hours ago
add a comment |
up vote
25
down vote
== is a binary equality operator.
The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First
answered 8 hours ago
Andrew Tobilko
24.4k84182
1
@JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for==. Without the parentheses, neither of these expressions would be syntactically correct.
– Andrew Tobilko
7 hours ago
add a comment |
up vote
25
down vote
up vote
25
down vote
== is a binary equality operator.
The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First
answered 8 hours ago
Andrew Tobilko
24.4k84182
== is a binary equality operator.
The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First
answered 8 hours ago
Andrew Tobilko
24.4k84182
edited 8 hours ago
answered 8 hours ago
Andrew Tobilko
24.4k84182
answered 8 hours ago
Andrew Tobilko
24.4k84182
answered 8 hours ago
Andrew Tobilko
24.4k84182
24.4k84182
1
@JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for==. Without the parentheses, neither of these expressions would be syntactically correct.
– Andrew Tobilko
7 hours ago
add a comment |
1
@JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for==. Without the parentheses, neither of these expressions would be syntactically correct.
– Andrew Tobilko
7 hours ago
1
1
@JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for
==. Without the parentheses, neither of these expressions would be syntactically correct.– Andrew Tobilko
7 hours ago
@JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for
==. Without the parentheses, neither of these expressions would be syntactically correct.– Andrew Tobilko
7 hours ago
add a comment |
up vote
15
down vote
As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.
//the example values
x = 1;
y = 2;
So in quickReplaceAndCompare the following is done:
x == (x = y)
1 == (x = y)
1 == (x = 2) //assign 2 to x, returns 2
1 == 2
false
and in badReplaceAndCompare():
(x = y) == x
(x = 2) == x //assign 2 to x, returns 2
2 == x
2 == 2
true
note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.
answered 8 hours ago
Poohl
33319
add a comment |
up vote
15
down vote
As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.
//the example values
x = 1;
y = 2;
So in quickReplaceAndCompare the following is done:
x == (x = y)
1 == (x = y)
1 == (x = 2) //assign 2 to x, returns 2
1 == 2
false
and in badReplaceAndCompare():
(x = y) == x
(x = 2) == x //assign 2 to x, returns 2
2 == x
2 == 2
true
note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.
answered 8 hours ago
Poohl
33319
add a comment |
up vote
15
down vote
up vote
15
down vote
As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.
//the example values
x = 1;
y = 2;
So in quickReplaceAndCompare the following is done:
x == (x = y)
1 == (x = y)
1 == (x = 2) //assign 2 to x, returns 2
1 == 2
false
and in badReplaceAndCompare():
(x = y) == x
(x = 2) == x //assign 2 to x, returns 2
2 == x
2 == 2
true
note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.
answered 8 hours ago
Poohl
33319
As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.
//the example values
x = 1;
y = 2;
So in quickReplaceAndCompare the following is done:
x == (x = y)
1 == (x = y)
1 == (x = 2) //assign 2 to x, returns 2
1 == 2
false
and in badReplaceAndCompare():
(x = y) == x
(x = 2) == x //assign 2 to x, returns 2
2 == x
2 == 2
true
note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.
answered 8 hours ago
Poohl
33319
answered 8 hours ago
Poohl
33319
answered 8 hours ago
Poohl
33319
answered 8 hours ago
Poohl
33319
33319
add a comment |
add a comment |
up vote
0
down vote
In the first test you're checking does 1 == 3.
In the second test your checking does 3 == 3.
(x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?
In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?
answered 39 mins ago
Michael Puckett II
3,67541436
add a comment |
up vote
0
down vote
In the first test you're checking does 1 == 3.
In the second test your checking does 3 == 3.
(x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?
In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?
answered 39 mins ago
Michael Puckett II
3,67541436
add a comment |
up vote
0
down vote
up vote
0
down vote
In the first test you're checking does 1 == 3.
In the second test your checking does 3 == 3.
(x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?
In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?
answered 39 mins ago
Michael Puckett II
3,67541436
In the first test you're checking does 1 == 3.
In the second test your checking does 3 == 3.
(x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?
In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?
answered 39 mins ago
Michael Puckett II
3,67541436
answered 39 mins ago
Michael Puckett II
3,67541436
answered 39 mins ago
Michael Puckett II
3,67541436
answered 39 mins ago
Michael Puckett II
3,67541436
3,67541436
add a comment |
add a comment |
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11
The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
9 hours ago
@LouisWasserman The term evaluate is, in my opinion, inapplicable here because
xdoes not need to be evaluated, it just loaded from memory.– John McClane
9 hours ago
3
Evaluating the expression
x = yis certainly relevant, and causes the side effect thatxis set to the value ofy.– Louis Wasserman
9 hours ago
@LouisWasserman Sure. I knew that, and it was the main idea that lead me to
quickReplaceAndCompare().– John McClane
9 hours ago
7
Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
6 hours ago