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这看起来很奇怪，我可以捕获静态变量，但前提是该变量未在捕获列表中指定，即它隐式捕获它。
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;
}
我不明白，第三次和第四次捕获应该是等价的，不是吗？在第三个中，我没有捕获具有“自动存储持续时间”的变量
编辑:我认为答案是它从不捕获静态变量，所以:
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.