Consider

#include <iostream>

struct Foo
{
    int* n;
    Foo(){n = new int{};}
    ~Foo(){delete n;}
    int& get()
    {
        int* m = n;
        return *m;
    } 
};

int main()
{
    Foo f;
    std::cout << f.get();
}

This is a cut-down version of a class that manages a pointer, and has a method that returns a reference to the dereferenced pointer.

Is that defined behaviour?

CodePudding user response：

Is that defined behaviour?

Yes, the given program is well-formed. You're returning a non-const lvalue reference that refers to a dynamically allocated integer pointed by the pointer n and m. The integer object still exists after the call f.get(). That is, itis not a function local variable.

Note also that just returning a reference to a potentially local variable is not undefined behavior in itself. It's just that if you were to use that returned reference(aka dangling reference) to a local variable that nolonger exists, then we will get UB.