Is the behaviour of i = i++ really undefined?

Question

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Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)

According to c++ standard,

i = 3; i = i++; 

will result in undefined behavior.

We use the term "undefined behavior" if it can lead to more then one result. But here, the final value of i will be 4 no matter what the order of evaluation, so shouldn't this really be called "unspecified behavior"?

Answer 1

The phrase, "…the final value of i will be 4 no matter what the order of evaluation…" is incorrect. The compiler could emit the equivalent of this:

i = 3; int tmp = i; ++i; i = tmp; 

or this:

i = 3; ++i; i = i - 1; 

or this:

i = 3; i = i; ++i; 

As to the definitions of terms, if the answer was guaranteed to be 4, that wouldn't be unspecified or undefined behavior, it would be defined behavior.

As it stands, it is undefined behaviour according to the standard (Wikipedia), so it's even free to do this:

i = 3; system("sudo rm -rf /"); // DO NOT TRY THIS AT HOME … OR AT WORK … OR ANYWHERE. 

Answer 2

No, we don't use the term "undefined behavior" when it can simply lead to more than one arithmetical result. When the behavior is limited to different arithmetical results (or, more generally, to some set of predictable results), it is typically referred to as unspecified behavior.

Undefined behavior means completely unpredictable and unlimited consequences, like formatting the hard drive on your computer or simply making your program to crash. And i = i++ is undefined behavior.

Where you got the idea that i should be 4 in this case is not clear. There's absolutely nothing in C++ language that would let you come to that conclusion.