C ++中lambda派生的隐式函子的生命周期是什么？ [英] What is lifetime of lambda-derived implicit functors in C++?

问题描述

问题很简单：当我写一个lambda表达式时，什么是C ++编译器为我自动生成的functor对象的生命周期？

The question is simple: what is lifetime of that functor object that is automatically generated for me by the C++ compiler when I write a lambda-expression?

一个快速搜索，但找不到一个满意的答案。特别是，如果我通过lambda某处，它被记住在那里，然后我走出范围，一旦我的lambda被调用后会发生什么，并试图访问我的堆栈分配，但不再活着，捕获的变量？还是编译器以某种方式防止这种情况？或者什么？

I did a quick search, but couldn't find a satisfactory answer. In particular, if I pass the lambda somewhere, and it gets remembered there, and then I go out of scope, what's going to happen once my lambda is called later and tries to access my stack-allocated, but no longer alive, captured variables? Or does the compiler prevent such situation in some way? Or what?

推荐答案

取决于如何捕获变量。如果你通过引用捕获它们（ [&] ），它们超出范围，引用将无效，就像正常引用一样。如果您想确保它们的作用范围，请按值（ [=] ）捕获它们。

Depends on how you capture your variables. If you capture them by reference ([&]) and they go out of scope, the references will be invalid, just like normal references. Capture them by value ([=]) if you want to make sure they outlife their scope.

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