Does Capturing a Reference by Reference in C 11 Lambdas Guarantee Access to the Modified Value?

Capturing a Reference by Reference in C 11 Lambdas

The Question:

Consider the following code snippet:

<code class="cpp">#include <functional>
#include <iostream>

std::function<void()> make_function(int&amp; x) {
    return [&amp;]{ std::cout << x << std::endl; };
}

int main() {
    int i = 3;
    auto f = make_function(i);
    i = 5;
    f();
}</code>

Can we guarantee that this program will output 5 without encountering undefined behavior? This question arises specifically when capturing the variable x by reference ([&]), and concerns whether capturing a reference to a variable will result in a dangling reference once the function make_function returns.

The Answer:

Yes, the code is guaranteed to work.

Explanation:

The C 11 lambda specification states that the reference captured here remains valid as long as the originally referenced object still exists. This means that even though the parameter x in make_function goes out of scope after the function returns, the lambda closure still retains a valid reference to the integer i.

Clarification:

To address some inaccuracies in previous responses:

• "Scope" in C refers to a static, lexical region of code where unqualified name lookup associates a name with a declaration. It has no direct relation to lifetime.
• "Reaching scope" rules for lambdas determine when capture is allowed based on syntactic rules, not lifetime.

In this specific case, the variable x is within the reaching scope of the lambda and is captured by reference. Therefore, the reference remains valid, and the lambda can continue to access the modified value of i.

Conclusion:

This code demonstrates the correct capture of a reference by reference in a lambda. It is guaranteed to output 5 without invoking undefined behavior.

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