Is this code:
y = x = x + 1;
undefined behavior in C?
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3 Answers
32
Answer to your question
No.
What will happen
This will happen:
int x = 1; /* ASSUME THIS IS SO */
y = x = x + 1;
/* Results: */
y == 2;
x == 2;
How it compiles
The same as:
x += 1;
y = x;
Why this is not undefined
Because you are not writing x in the same expression you read it. You just set it to itself + 1, then assign y to the value of x.
Your future
If you find the code confusing you can use parentheses for readability:
y = x = (x + 1);
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2Just for your info, muntoo, as the "return value" of assignment is it's own value, you can also use it for other things, like `if ( 1 == x = 1 )`, which is a pretty common bug and the reason, developers prefer to write `1 == x` instead of `x == 1` (because you can't assign x to 1). – anroesti Feb 05 '11 at 23:02
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34The only developers I know who write things like `1 == x` are ones I would not work with... – R.. GitHub STOP HELPING ICE Feb 05 '11 at 23:21
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19@andre: No, "developers" don't prefer to write that. It's gross, unclean, and useless; just join the last decade and turn on your compiler warnings. – GManNickG Feb 05 '11 at 23:43
17
No, your expression is properly defined. You probably were looking for y = x = x++;, which is not.
Nikolai Fetissov
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Wouldn't `y = x = x++;` be equal to `y = x = x; x += 1;`? Why wouldn't that be properly defined? – rzetterberg Jul 10 '11 at 08:37
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Or would it be equivalent to `y = x = x + 1; x += 1;`, or to `y = x + 1 = x; x += 1;`? – Nikolai Fetissov Jul 10 '11 at 12:57
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I was asking because I wasn't sure and I wanted to know why `x + 1` was ok, but `x++` wasn't. – rzetterberg Jul 10 '11 at 13:16
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1That example statement is supposed to change the value of `x` twice. The problem is that the C++ language does not tell in what order that would happen. – Nikolai Fetissov Jul 10 '11 at 16:16
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9
No. You only modify x once, and due to the right-associativity of = that assignment happens before the assignment to y. Even if it did happen after, there's still only one modification of x. Your statement is as legal as y = ++x.
Michael Mrozek
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Chris Lutz
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