Why rvalue reference binding to xvalue doesn't work in my code?

Question

I tried to understand lvalue and rvalue in C++11. So I wrote a test code:

int x = 10;
int foo() { return x; }
int& bar() { return x; }
int&& baz() { return 10; }

int main() {
    int& lr1 = 10;     // error: lvalue references rvalue
    int& lr2 = x;      // ok
    int& lr3 = foo();  // error: lvalue references rvalue
    int& lr4 = bar();  // ok
    int& lr5 = baz();  // error: lvalue references rvalue

    int&& rr1 = 10;    // ok
    int&& rr2 = x;     // error: rvalue references lvalue
    int&& rr3 = foo(); // ok
    int&& rr4 = bar(); // error: rvalue references lvalue
    int&& rr5 = baz(); // ok
}

It works pretty well, so I inserted std::cout to print results.

#include <iostream>

int x= 10;
int foo() { return x; }
int& bar() { return x; }
int&& baz() { return 10; }

int main() {
    int& lr1 = 10;     std::cout << lr1 << std::endl; // error
    int& lr2 = x;      std::cout << lr2 << std::endl; // ok
    int& lr3 = foo();  std::cout << lr3 << std::endl; // error
    int& lr4 = bar();  std::cout << lr4 << std::endl; // ok
    int& lr5 = baz();  std::cout << lr5 << std::endl; // error

    int&& rr1 = 10;    std::cout << rr1 << std::endl; // ok
    int&& rr2 = x;     std::cout << rr2 << std::endl; // error
    int&& rr3 = foo(); std::cout << rr3 << std::endl; // ok
    int&& rr4 = bar(); std::cout << rr4 << std::endl; // error
    int&& rr5 = baz(); std::cout << rr5 << std::endl; // ERROR!?
}

int&& rr5 = baz(); std::cout << rr5; causes a Runtime Error, but I don't know why it makes an error.

I think the return value of baz() would be xvalue, so its lifetime is prolonged. But when I tried to access its value, the error occurs. Why?

Answer 1

I think the return value of baz() would be xvalue, so its lifetime is prolonged.

At first what baz() returns is always a dangling reference.

For int&& baz() { return 10; } int&& baz() { return 10; } , the lifetime of the temporary is not extended. It's constructed inside the function and will be destroyed when get out of the function, then baz() always returns a dangling reference.

a temporary bound to a return value of a function in a return statement is not extended: it is destroyed immediately at the end of the return expression. Such function always returns a dangling reference.

Then for int&& rr5 = baz(); , rr5 is a dangling reference too; deference on it leads to UB and anything is possible.

On the other hand, if you change baz() to return-by-value, everything would be fine; the return value is copied and then bound to rr5 , then the lifetime of the temporary is extended to the lifetime of rr5 .

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