Returning reference to temporary (built in types)

Question

From this thread it is clear that a string literal cannot be used for a function that returns const string& since there is an implicit conversion from const char* to std::string which creates a temporary.

But then why do I get a warning "warning: returning reference to a temporary" if my return type match exactly and there is no need for conversion, e.g.:

include <iostream>

const int& test(){
    return 2;
}

int main(){

    std::cout << test();

}

No implicit conversion needed to happen on the return value of 2, so why is there a warning? I thought using test() would be pretty much the same as doing

 const int& example = 2;

which is perfectly valid. Additionally if I change 2 to 2.2 (so it is a double) the program still runs (with the same warning) despite the fact there IS a conversion from double to int? Shouldn't I be running in to an issue similar to how const char* was returned to the string reference, if there is a conversion from double to int?

Answer 1

A temporary is still created. §8.5.3/(5.2.2.2) applies¹:

Otherwise, a temporary of type “ cv1 T1” is created and copy-initialized (8.5) from the initializer expression. The reference is then bound to the temporary.

This also applies in your second example. It does not apply for prvalues of class type, or scalar xvalues: Both

const A& a = A();
// and
const int& i = std::move(myint);

do not introduce a temporary. However, that isn't changing the final result : In any case, the temporary that is bound to the reference will be destroyed at the end of the return statement - §12.2/(5.2):

The lifetime of a temporary bound to the returned value in a function return statement (6.6.3) is not extended; the temporary is destroyed at the end of the full-expression in the return statement.

That is, the temporary is destroyed before the function even exits, and thus your program induces undefined behavior.

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¹ I could go on and quote the entire list to show why it does, but that would presumably be a waste of answer space.