CODEWITHSUNDEEP
c++
Rick
Rick

Reputation: 7506

What is the type of decltype(*int_iterator + 0)?

template <typename AIterator>
auto foo(AIterator begin) -> decltype(*begin + 0){
    return *begin;
}

For example:

vector<int> ivec = {1,2,3};
foo(ivec.begin());

My answer book says it's a const reference type, is it true?

But IIRC, decltype(int reference + int), expression type is a int (*begin is a reference, so naturally I think *begin + 0 should also result in a int).

For example:

int a = 3, &b = a;
decltype(b + 0) d; //d is a int

PS: I tried on VS and the IDE hints the return type of function foo is int.

The book is C++ Primer 5th answer book, but mine is not a English version and it's adapted by the translators, so I didn't mention it at first.

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Upvotes: 1

Views: 211

Answers (1)

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385164

It's int, and this is exactly why the author has added + 0 … as a quick way to get rid of the reference type and ensure that foo returns a copy (which appears to be its purpose).

decltype on a prvalue expression of type T always evaluates to T.

Either the book's prose is wrong, or you've misread it.

Upvotes: 3

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