c++ - Lambda 函数捕获变量与返回值？

Question

我正在学习 C++11 新函数 lambda 函数，有点困惑。 我看过了

 []         Capture nothing (or, a scorched earth strategy?)
 [&]        Capture any referenced variable by reference
 [=]        Capture any referenced variable by making a copy
 [=, &foo]  Capture any referenced variable by making a copy, but capture variable foo by reference
 [bar]      Capture bar by making a copy; don't copy anything else
 [this]     Capture the this pointer of the enclosing class

我的问题是捕获变量到底是什么意思，返回值有什么区别，如果你想捕获一个变量，你必须正确地返回它？

例如:

[] () { int a = 2; return a; } 

为什么不是

int [] () { int a = 2; return a; }

Answer 1

您可以“捕获”属于封闭函数的变量。

void f() {
    int a=1;
    // This lambda is constructed with a copy of a, at value 1.
    auto l1 = [a] { return a; };
    // This lambda contains a reference to a, so its a is the same as f's.
    auto l2 = [&a] { return a; };

    a = 2;

    std::cout << l1() << "\n"; // prints 1
    std::cout << l2() << "\n"; // prints 2
}

如果需要，您可以返回捕获的变量，也可以返回其他内容。返回值与捕获无关。