C ++ 11 lambda作为成员变量？ [英] C++11 lambda as member variable?

问题描述

可以将lambda定义为类成员吗？

Can lambda's be defined as class members?

例如，是否可以使用lambda而不是函数对象重写下面的代码示例？ p>

For example, would it be possible to rewrite the code sample below using a lambda instead of a function object?

struct Foo {
    std::function<void()> bar;
};

我不知道的原因是因为以下lambda可以作为参数传递：

The reason I wonder is because the following lambda's can be passed as arguments:

template<typename Lambda>
void call_lambda(Lambda lambda) // what is the exact type here?
{ 
    lambda();
}

int test_foo() {
    call_lambda([]() { std::cout << "lambda calling" << std::endl; });
}

我想如果一个lambda可以作为一个函数参数传递，

I figured that if a lambda can be passed as a function argument then maybe they can also be stored as a member variable.

更多的修补后，我发现这是有效的（但它是无意义的）：

After more tinkering I found that this works (but it's kind of pointless):

auto say_hello = [](){ std::cout << "Hello"; };
struct Foo {
    typedef decltype(say_hello) Bar;
    Bar bar;
    Foo() : bar(say_hello) {}
};

推荐答案

模板可以没有类型擦除， ：

Templates make it possible without type erasure, but that's it:

template<typename T>
struct foo {
    T t;
};

template<typename T>
foo<typename std::decay<T>::type>
make_foo(T&& t)
{
    return { std::forward<T>(t) };
}

// ...
auto f = make_foo([] { return 42; });

重复每个人都已经暴露的参数： [] {} code>不是一个类型，所以你不能使用它作为eg您正在尝试的模板参数。使用 decltype 也是iffy，因为lambda表达式的每个实例都是具有唯一类型的单独的闭包对象的符号。 （例如，上面的 f 的类型不是 foo< decltype（[] {return 42;}）> / code>。）

Repeating the arguments that everyone has already exposed: []{} is not a type, so you can't use it as e.g. a template parameter like you're trying. Using decltype is also iffy because every instance of a lambda expression is a notation for a separate closure object with a unique type. (e.g. the type of f above is not foo<decltype([] { return 42; })>.)

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