C ++ lambda表达式-编译器如何解释它们？ [英] C++ lambda expressions - How does the compiler interpret them?

问题描述

我刚刚开始学习C ++ 11的新功能。我正在阅读有关C ++ Primer（Stanley Lippman）中的lambda的文章，并正在尝试它们。

I just starter learning the new features in C++ 11. I was reading about lambdas in C++ Primer (Stanley Lippman) and was experimenting with them.

我尝试了以下代码段：

auto func() -> int (*) (){
    //int c=0;
    return []()-> int {return 0;};
}

int main(){
    auto p = func();
}

此代码编译良好。因此，我猜没有任何捕获的lambda只是由编译器作为普通函数生成的，我们可以使用指向它们的普通函数指针。

This code compiled fine. So I guess lambdas without any captures are just generated as normal functions by the compiler and we can use a normal function pointer to them.

现在，我更改了代码以使用捕获：

Now I changed the code to use captures:

auto func() -> int (*) (){
    int c=0;
    return [=]()-> int {return c;};
}

int main(){
    auto p = func();
}

但这无法编译。使用g ++时出现以下编译错误：

But this failed to compile. I got the following compilation error while using g++:

main.cpp: In function ‘int (* func())()’:
main.cpp:6:31: error: cannot convert ‘func()::__lambda0’ to ‘int (*)()’ in return
return [=]()-> int {return c;};

从错误中我可以理解，它不是生成的普通函数，可能是具有重载的呼叫操作符的类。还是其他东西？

From the error I can understand that it is not a normal function that is generated and it might probably be a class with an overloaded call-operator. Or is it something else?

我的问题：编译器如何内部处理lambda？我该如何传递使用捕获的lambda，即func（）的返回值应该是什么？我目前无法想到一个用例，在该用例中，我需要使用像这样的lambda，但我只想了解更多有关它们的信息。请帮忙。

My questions : How does the compiler handle lambdas internally? How should I pass around lambdas that use captures i.e. what should be the return value from func()? I cant currently think of a use case where I would need to use lambdas like this but I just want to understand more about them. Please help.

谢谢。

推荐答案

所有lambda都是函数对象具有定义为 closure type 且具有 operator（）成员的实现的类型。每个lambda表达式也都有自己独特的闭包类型。

All lambdas are function objects with implementation defined type called closure type with a operator() member. Every lambda expression has it's own unique closure type, too.

没有捕获的Lambda可以转换为。

Lambdas without a capture can be converted to a function pointer. Whether or not compiler generates a normal function behind the scenes is an internal detail and shouldn't matter to you.

不可能返回在函数内部定义的lambda，这是内部细节，对您而言无关紧要。 。几乎没有什么可以防止这种情况的发生-您不知道lambda表达式类型的名称，不能在 decltype 内使用lambda表达式，正如已经说过的，两个lambda表达式（即使在词法上相同）也具有不同的类型。

It's not possible to return a lambda that's defined inside a function. There are few things that prevent this - you don't know the name of a type of lambda expression, you can't use a lambda expression inside a decltype and as already said, two lambda expressions (even if lexically identical) have different types.

您可以使用 std :: function ：

std::function<int()> func()
{
    int i = 0;
    return [=]()-> int {return i;};
}

这种方式也适用于捕获。

This way works with captures, too.

或类似这样的东西：

auto f = []{ return 0; };

auto func() -> decltype(f)
{
    return f;
}

编辑：即将推出的C ++ 1y标准（更具体地说，返回类型推导），但可以执行以下操作：

The upcoming C++1y standard (more specifically, return type deduction), however, will allow you to do this:

auto func()
{
    int i = 42;
    return [=]{ return i; };   
}

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