In the accepted answer to "Iterator to last element of std::vector using end()--" @barry states:

Note that if vector::iterator is just T* (which would be valid), the first form above is ill-formed. The second two work regardless, so are preferable.

referring to his code:

std::vector<int>::iterator it = --container.end();
std::vector<int>::iterator it = container.end() - 1;
std::vector<int>::iterator it = std::prev(container.end());

This opinion is disputed in the comments, however without a clear resolution. So that's my question: what exactly is the semantic difference between the first and the second? And would the answer be different for iterators over structures other than vector?

CodePudding user response：

For any standard library container, the member function end() returns an r-value. It's a "temporary" until you assign it to a variable.

The decrement operator -- is not required to work on r-value iterators. You would be modifying a temporary, which C historically has taken measures to avoid.

Therefore, --container.end() might compile on your standard-conforming C compiler. But it might not.

std::prev(container.end()) will work on every standard-conforming compiler.

To review:

• --container.end() may not compile. It is up to the implementation.
• container.end() - 1 will only compile if the container uses random-access iterators.
• std::prev(container.end()) will always compile.

All three forms will produce the same result, if they compile.