无法在Lambda中捕获静态变量 [英] Can't capture a static variable in a lambda

问题描述

这似乎很奇怪，我可以捕获静态变量，但前提是在捕获列表中未指定该变量，即隐式捕获它。

This seems strange, I can capture the static variable, but only if the variable isn't specified in the capture list, i.e., it implicitly captures it.

int main()
{
    int captureMe = 0;
    static int captureMe_static = 0;

    auto lambda1 = [&]() { captureMe++; };  // Works, deduced capture
    auto lambda2 = [&captureMe]() { captureMe++; }; // Works, explicit capture
    auto lambda3 = [&] () { captureMe_static++; };  // Works, capturing static int implicitly
    auto lambda4 = [&captureMe_static] { captureMe_static++; }; // Capturing static in explicitly:  
                                                                // Error: A variable with static storage duration 
                                                                // cannot be captured in a lambda

    // Also says "identifier in capture must be a variable with automatic storage duration declared
    // in the reaching scope of the lambda

    lambda1(); lambda2(); lambda3();    // All work fine

    return 0;
}

我不理解，第三个和第四次捕获应该是等效的，不是吗？在第三次中，我不是捕获具有自动存储持续时间的变量

I'm not understanding, the third and fourth captures should be equivalent, no? In the third I'm not capturing a variable with "automatic storage duration"

编辑：我认为答案是这样的永远不会捕获静态变量，所以：

I think the answer to this is that it is never capturing the static variable, so:

auto lambda = [&] { captureMe_static++; };   // Ampersand says to capture any variables, but it doesn't need to capture anything so the ampersand is not doing anything
auto lambda = [] { captureMe_static++; };  // As shown by this, the static doesn't need to be captured, and can't be captured according to the rules.

推荐答案

具有静态存储持续时间的变量不需要被捕获，因此无法被捕获。您可以在lambda中简单地使用它。

A variable with static storage duration doesn't need to be captured, and, therefore, cannot be captured. You can simply use it within the lambda.

对于自动变量，存在一个问题：而在其他语言中，闭包仅将对变量的引用存储在封闭范围内，在C ++中，lambda无法延长自动变量的生存期，因此可能超出范围，从而在lambda中留下了悬挂的引用。因此，C ++允许您选择是通过复制还是通过引用捕获自动变量。但是，如果变量是静态的，则不会出现此问题。该lambda的行为就像是通过引用捕获它一样。

With automatic variables there is an issue: while in other languages a closure would simply store a reference to the variable in the enclosing scope, in C++ a lambda does not have the ability to extend the lifetime of an automatic variable, which might therefore go out of scope, leaving behind a dangling reference in the lambda. For this reason C++ allows you to choose whether to capture an automatic variable by copy or by reference. But if the variable is static then this issue does not arise; the lambda will simply behave as though it has captured it by reference.

如果您确实想通过 value 捕获静态变量，请使用C ++ 14初始化捕获语法。

If you do want to capture a static variable by value then use the C++14 init-capture syntax.

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