Why does the following closure not work?

Question

Consider the following code.

#include <functional>
#include <stdio.h>
#include <stdlib.h>


std::function<void()> getFunction() {
    int foo = 0;
    return [&] () {
        printf("foo = %d\n", foo);
    };
}
int main(){
    std::function<void()> foo = getFunction();
    foo();
}

On my machine, it prints the following.

foo = 32767

Why does it not print 0?

possible duplicate of [In C++11 Lambdas, what happens if I capture a local variable by reference, and it goes out of scope?](http://stackoverflow.com/questions/9488129/in-c11-lambdas-what-happens-if-i-capture-a-local-variable-by-reference-and-i). [This](http://stackoverflow.com/questions/12463269/c11-lambda-closure-involving-a-stack-variable-by-reference-that-leaves-scope-i) too. — user657267, Nov 04 '14 at 01:06

score 0 · Answer 1 · answered Nov 04 '14 at 01:02

It turns out, after consulting one of the other students in my lab, that if you want to capture a variable in a closure by reference, you must ensure that the said variable is still in scope at the time you try to access it. In particular, this means that you can pass a closure to a function call from the function getFunction() but not return it to another function. The closure above has an equivalent problem to returning a local variable by pointer.

Instead, one can use the value-capturing closure, although that means that the captured variables are read-only.

std::function<void()> getFunction() {
    int foo = 0;
    return [=] () { // Capture by value works
        printf("foo = %d\n", foo);
    };
}

`although that means that the captured variables are read-only` you can make the lambda `mutable` if you want to modify copy-captured entities. Then again all of this has been explained multiple times across other questions, did you try searching first? — user657267, Nov 04 '14 at 01:09

Why does the following closure not work?

1 Answers1