c++ - Lambda 捕获 C++14

Question

我遇到过这样的符号:

int x = 4;
auto y = [&r = x, x = x+1]()->int { 
    r += 2;
    return x+2;
}();

你能解释一下这个说法吗？我是 C++03 的用户，最近升级到 C++11。从今天开始，我开始使用 C++14 并遇到了这个片段。

谢谢!

Answer 1

感谢@chris wikipedia reference .我发现的是 -

Here is nice explanation who don't know about the old lambda Captures of C++11

在 C++14 中:

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C++11 lambda functions capture variables declared in their outer scope by value-copy or by reference. This means that value members of a lambda cannot be move-only types. C++14 allows captured members to be initialized with arbitrary expressions. This allows both capture by value-move and declaring arbitrary members of the lambda, without having a correspondingly named variable in an outer scope.

This is done via the use of an initializer expression:

auto lambda = [value = 1] {return value;};

The lambda function lambda will return 1, which is what value was initialized with. The declared capture deduces the type from the initializer expression as if by auto.

This can be used to capture by move, via the use of the standard std::move function:

std::unique_ptr<int> ptr(new int(10));
auto lambda = [value = std::move(ptr)] {return *value;};

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因此上面的表达式将 x 更新为 6，并将 y 初始化为 7。