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Home/ Questions/Q 148345
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Editorial Team
Asked: May 11, 20262026-05-11T08:56:31+00:00 2026-05-11T08:56:31+00:00

There’s been some debate going on in this question about whether the following code

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There’s been some debate going on in this question about whether the following code is legal C++:

std::list<item*>::iterator i = items.begin(); while (i != items.end()) {     bool isActive = (*i)->update();     if (!isActive)     {         items.erase(i++);  // *** Is this undefined behavior? ***     }     else     {         other_code_involving(*i);         ++i;     } } 

The problem here is that erase() will invalidate the iterator in question. If that happens before i++ is evaluated, then incrementing i like that is technically undefined behavior, even if it appears to work with a particular compiler. One side of the debate says that all function arguments are fully evaluated before the function is called. The other side says, ‘the only guarantees are that i++ will happen before the next statement and after i++ is used. Whether that is before erase(i++) is invoked or afterwards is compiler dependent.’

I opened this question to hopefully settle that debate.

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1 Answer

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  1. 2026-05-11T08:56:32+00:00Added an answer on May 11, 2026 at 8:56 am

    Quoth the C++ standard 1.9.16:

    When calling a function (whether or not the function is inline), every value computation and side effect associated with any argument expression, or with the postfix expression designating the called function, is sequenced before execution of every expression or statement in the body of the called function. (Note: Value computations and side effects associated with the different argument expressions are unsequenced.)

    So it would seem to me that this code:

    foo(i++); 

    is perfectly legal. It will increment i and then call foo with the previous value of i. However, this code:

    foo(i++, i++); 

    yields undefined behavior because paragraph 1.9.16 also says:

    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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