This code snippet:
int& func1()
{
int i;
i = 1;
return i;
}
will not work because you're returning an alias (a reference) to an object with a lifetime limited to the scope of the function call. That means once func1() returns, int i dies, making the reference returned from the function worthless because it now refers to an object that doesn't exist.
int main()
{
int& p = func1();
/* p is garbage */
}
The second version does work because the variable is allocated on the free store, which is not bound to the lifetime of the function call. However, you are responsible for deleteing the allocated int.
int* func2()
{
int* p;
p = new int;
*p = 1;
return p;
}
int main()
{
int* p = func2();
/* pointee still exists */
delete p; // get rid of it
}
Typically you would wrap the pointer in some RAII class and/or a factory function so you don't have to delete it yourself.
In either case, you can just return the value itself (although I realize the example you provided was probably contrived):
int func3()
{
return 1;
}
int main()
{
int v = func3();
// do whatever you want with the returned value
}
Note that it's perfectly fine to return big objects the same way func3() returns primitive values because just about every compiler nowadays implements some form of return value optimization:
class big_object
{
public:
big_object(/* constructor arguments */);
~big_object();
big_object(const big_object& rhs);
big_object& operator=(const big_object& rhs);
/* public methods */
private:
/* data members */
};
big_object func4()
{
return big_object(/* constructor arguments */);
}
int main()
{
// no copy is actually made, if your compiler supports RVO
big_object o = func4();
}
Interestingly, binding a temporary to a const reference is perfectly legal C++.
int main()
{
// This works! The returned temporary will last as long as the reference exists
const big_object& o = func4();
// This does *not* work! It's not legal C++ because reference is not const.
// big_object& o = func4();
}