Incremented iterator std::next when swapping

Question

Consider the code:

list<int> a{ 4,3,1,2 }; auto i = a.begin(); swap(*i, *(++i));

Why does the swap do nothing? While the following works as expected?

list<int> a{ 4,3,1,2 }; auto i = a.begin(); swap(*i, *(next(i)));

@eerorika the evaluation order of parameters is very much compiler dependent, so they may get different results than you do. — Mark Ransom
– Mark Ransom, Commented Aug 17, 2020 at 15:22
There is a simpler way to swap: std::iter_swap(i, std::next(i)) — eerorika
– eerorika, Commented Aug 17, 2020 at 15:29

asmmo · Accepted Answer · 2020-08-17 15:29:54Z

In the first code, the order of evaluation of the two ops *i and *(++i) unspecified after c++17, hence the second one may execute before the first, and then the swapping swaps two equivalent values.

In the attached link, u can see that

f(++i, ++i); // undefined behavior until C++17, unspecified after C++17

But in the second code you have different parameters and std::next() returns a new iterator.

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