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Asked: 2026-05-26T09:22:36+00:00

According to FAQ , i = i++ is a undefined behaviour in C because

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According to FAQ, i = i++ is a undefined behaviour in C because this statement has only one sequence point (full expression), and in this statement i has been changed twice (side effect of i++ and =) so it is an undefined behaviour.

Question 1 is do I get that correctly? Or do I misunderstand why i = i++ is undefined?

The second question is that is x = i++ an valid expression?

I guess it is valid and the value of x will always be the origin value of i. Because although there is only one sequence point in this statement, but both x and i are modified only once, and i++ has a higher precedence, which means it should be valid and x++ will always done before the assignment, make x equals to the origin value of x. Is that correct?

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1 Answer

  1. Editorial Team
    1. Yes, at least as of C++03. I believe C++11 changes this somewhat, but I can’t get my hands on a copy of that standard to check.
    2. Because you modify x once, and i once. There’s no multiple writes to a single variable without an intervening sequence point.
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