Question

template <typename dataTypeA, 
          typename dataTypeB> 
                             dataTypeB const& functionX (dataTypeA argA, 
                                                         dataTypeB const& argB)
{
    return argA;
}

int main ()
{
    cout << functionX (3, 1L);
    return 0;
}

The compilation:

anisha@linux-dopx:~/Desktop/notes/c++> g++ functionTemplates.cpp -Wall -Wextra -pedantic

functionTemplates.cpp: In function ‘const dataTypeB& functionX(dataTypeA, const dataTypeB&) [with dataTypeA = int, dataTypeB = long int]’:
functionTemplates.cpp:47:26:   instantiated from here
functionTemplates.cpp:35:9: warning: returning reference to temporary

and then:

anisha@linux-dopx:~/Desktop/notes/c++> ./a.out
3

Why is it returning 3?

Isn’t argA a local variable for that function? Returning its reference shouldn’t be successful, isn’t it?

Answer 1

The compiler issues a warning, that you are returning an reference to local variable.

It works because returning a reference to local variable from a function is Undefined Behavior.
Undefined Behavior means anything can happen and the behavior cannot be explained within the semantics of the C++ Standard.

You are just being lucky, rather unlucky that it works. It may not work always.