CODEWITHSUNDEEP
c++c++11lambdareference
Li Chen
Li Chen

Reputation: 5260

output reference-passed value in lambda from(std::function)?

#include "stdafx.h"
#include <functional>
#include <iostream>
#include <string>
std::function<void(int)> Foo()
{
    int v = 1;
    int r = 2;
    auto l = [v, &r](int i)
    {
        std::cout << v << " " << r << " " << i << std::endl;
    };
    return l;
}

int main()
{
    auto func = Foo();
    func(3);
    return 0;
}

I think will print "1 2 3", however, it is "1 -858993460 3", why?

enter image description here

Upvotes: 1

Views: 57

Answers (1)

songyuanyao
songyuanyao

Reputation: 172864

Because r is captured by reference, but it's a local variable and will be destroyed immediately after Foo() returns. For func(3); the captured reference has become invalid, and deference on it leads to UB; anything is possible.

On the other hand, v is captured by value, that means it will be copied into the lambda and then works well.

Upvotes: 2

Related Questions