C++11 lambda 返回 lambda

问题描述

这段代码对于JS开发者来说并不是陌生的

this piece of code is not something unknown to JS developers

function get_counter()
{
    return (
        function() {
            var c = 0;
            return function() { return ++c; };
        })();
}

它基本上创建了一个，它创建了不同的枚举器.所以我想知道是否可以使用新的 lambda 语义在 C++11 中完成同样的事情?我最终写了这段 C++ 不幸的是不能编译！

it basically creates a which creates different enumerators. So I was wondering if same thing can be done in C++11 with new lambda semantics? I ended up writing this piece of C++ which unfortunately does not compile!

int main()
{
    int c;
    auto a = [](){
        int c = 0;
        return [&](){
            cout << c++;
        };
    };
    return 0;
}

所以我想知道是否有解决方法可以编译它，如果有编译器如何使此代码正确运行?我的意思是它必须创建单独的枚举器，但它也应该收集垃圾(未使用的 c 变量).

so I was wondering if there is a workaround to get it compiled and if there is how can compiler make this code run correctly? I mean it has to create separate enumerators but it should also collect garbage (unused c variables).

顺便说一下，我使用的是 VS2012 编译器，它会生成此错误:

by the way I'm using VS2012 compiler and it generates this error:

Error   2   error C2440: 'return' : cannot convert from 'main::<lambda_10d109c73135f5c106ecbfa8ff6f4b6b>::()::<lambda_019decbc8d6cd29488ffec96883efe2a>' to 'void (__cdecl *)(void)'    c:usersalidocumentsvisual studio 2012projects	est	estmain.cpp   25  1   Test

推荐答案

您的代码有一个错误，它包含一个悬空引用；c 引用将引用外部 lambda 中的局部变量，当外部 lambda 返回时该变量将被销毁.

Your code has a bug in that it contains a dangling reference; the c reference will refer to the local variable in the outer lambda, which will be destroyed when the outer lambda returns.

您应该使用 mutable 按值 lambda 捕获来编写它:

You should write it using a mutable by-value lambda capture:

auto a = []() {
    int c = 0;
    return [=]() mutable {
        cout << c++;
    };
};

这依赖于后标准扩展，以允许在返回类型推导的 lambda 中使用多个语句；如果 lambda 包含多个语句，是否有理由不允许它推导出返回类型? 最简单的修复方法是提供一个参数，以便 lambda 只包含一个语句:

This relies on a post-standard extension to allow multiple statements in a return-type-deducing lambda; Is there a reason on not allowing lambdas to deduce the return type if it contains more than one statement? The easiest way to fix it is to supply a parameter so that the lambda contains only a single statement:

auto a = [](int c) {
    return [=]() mutable {
        cout << c++;
    };
};

不幸的是，lambdas 中不允许使用默认参数，因此您必须将其称为 a(0).或者，以可读性为代价，您可以使用嵌套的 lambda 调用:

Unfortunately default parameters aren't allowed in lambdas, so you'd have to call this as a(0). Alternatively at the cost of readability you could use a nested lambda call:

auto a = []() {
    return ([](int c) {
        return [=]() mutable {
            cout << c++;
        };
    })(0);
};

这种工作方式是，当 a 执行内部 lambda 时，会将所有引用的变量复制到其闭包类型的一个实例中，如下所示:

The way this works is that when a executes the inner lambda copies all the referenced variables into an instance of its closure type, which here would be something like:

struct inner_lambda {
    int c;
    void operator()() { cout << c++; }
};

闭包类型的实例然后由外部 lambda 返回，并且可以被调用并在调用时修改它的 c 副本.

The instance of the closure type is then returned by the outer lambda, and can be invoked and will modify its copy of c when called.

总的来说，你的(固定的)代码被翻译成:

Overall, your (fixed) code is translated to:

struct outer_lambda {
    // no closure
    struct inner_lambda {
        int c;    // by-value capture
        // non-const because "mutable"
        void operator()() { cout << c++; }
    }
    // const because non-"mutable"
    inner_lambda operator()(int c) const {
        return inner_lambda{c};
    }
};

如果您将 c 作为按引用捕获保留，则为:

If you left c as a by-reference capture, this would be:

struct outer_lambda {
    // no closure
    struct inner_lambda {
        int &c;    // by-reference capture
        void operator()() const { cout << c++; } // const, but can modify c
    }
    inner_lambda operator()(int c) const {
        return inner_lambda{c};
    }
};

这里的 inner_lambda::c 是对局部参数变量 c 的悬空引用.

Here inner_lambda::c is a dangling reference to the local parameter variable c.

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