about reference in C++

Question

#include <iostream>
#include <string>

using namespace std;


string& funRef (string& s) {
string t = s;


return t;
}

int main() {
string s1 = "I'm a string!";
string& s2 = funRef(s1);
s2 = "YOUR STRING IS MINE";

cout << s1 << endl;
cout << s2 << endl;
}

why does it print out:

I'm a string!

\367\277_\377\367\277_\377M

I thought s2 was assigned with "YOUR STRING IS MINE".

what happen to the s2?

It references a destroyed object, `string t = s;` `t` is destroyed when `funref` is no longer in scope. — George, Feb 06 '17 at 23:43

score 0 · Answer 1 · answered Feb 06 '17 at 23:43

Most of your code is irrelevant.

This comes down to simple undefined behaviour, because here you are returning a reference to a local variable:

string& funRef(string& s)
{
   string t = s;
   return t;
}

The string t dies as soon as the call to funRef ends, but you're trying to use a reference to t later.

You can't do that.

about reference in C++

1 Answers1