Please help me understand the ouput.

Here is the code,

Code:

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int i = 1;
i = i++;

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At the end, i is still 1 instead of 2,why? Thanks.

Undefined behavior... look up "sequence points"

http://en.wikipedia.org/wiki/Sequence_point

See 2nd paragraph.

Quote:

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Originally Posted by Philip Nicoletti

Undefined behavior... look up "sequence points"

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I have a feeling it's defined in C++ 11. At least that's how I interpret section 5.2.6 in the standard: "The value computation of the ++ expression is sequenced before the modification of the operand object". In that case i = i++ is equivalent to this,

Code:

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int eval = i; // i++ is evaluated before incrementation
i++; // i is incremented
i = eval // the assignment is performed

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But it's still bad programming. If an outcome surprises you it probably surprises most programmers so it should be avoided.

Quote:

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Originally Posted by nuzzle

At least that's how I interpret section 5.2.6 in the standard: "The value computation of the ++ expression is sequenced before the modification of the operand object".

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I think that just establishes the "post-increment/decrement" aspect, i.e., the value of i++ is guaranteed to be the value of i since that is sequenced before the change to i. It does not mean that the increment operation is sequenced before the assignment operation in the code in post #1.

In other words, I think that the wording would still allow a conforming compiler to translate it as if it were:

Code:

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int eval = i; // i++ is evaluated before incrementation
i = eval // the assignment is performed
i++; // i is incremented

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Quote:

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Originally Posted by laserlight

In other words, I think that the wording would still allow a conforming compiler to translate it as if it were:

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Well, it looks like you're right. VS2012 actually outputs 2.

Heh, thanks for the confirmation, but I just remembered: we still have a similiar rule in place, so the behaviour is not just implementation defined in C++11, but still undefined as in C++03:

Quote:

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Originally Posted by C++11 Clause 1.9 Paragraph 15 (part)

If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.

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