CODEWITHSUNDEEP
c++
omzy
omzy

Reputation: 93

C++ reference to local variable that went out of scope

Very simple question, which most likely calls for an explanation on how references work and why my understanding is flawed.

Given the simple code snippet below: 

#include <iostream>
#include <string>

struct Foo
{
    std::string &ref;
    Foo(std::string &bar) : ref(bar) { }
};

int main()
{
    std::string s1 = "foo1";
    Foo f(s1);
    f.ref[0] = 'b';
    std::cout << s1 << std::endl; // prints boo1
    {
        std::string f2("tmp");
        f.ref = f2;
    } 
    // shouldn't f.ref be dangling by now?
    std::cout << f.ref; // prints tmp
}

Output:

My understanding is that f2 will be destroyed at the end of that block, thus f.ref will be a dangling reference. What is really going on?

Upvotes: 1

Views: 95

Answers (1)

M.M
M.M

Reputation: 141574

f.ref = f2 is the same as s1 = s2, because f.ref and s1 are both names for the same string object.

Perhaps you are mixing up the concepts of initialization of a reference, and assignment of objects. A reference can only be initialized once (and in fact, must be initialized on creation).

In the initialization ref(bar), ref is initialized to refer to the same object bar refers to (which is also the same object as is called s1).

In the assignment expression f.ref = f2; the assignment operator is called for the object which f.ref names.

Upvotes: 3

Related Questions