比较std :: function和lambda [英] Compare of std::function with lambda

问题描述

如何将std :: function与lambda进行比较?

How to compare of std::function with lambda?

#include <iostream>
#include <functional>
using namespace std;

int main()
{
    using Type = void( float, int );
    std::function<Type> a;
    auto callback = []( float d, int r )
    {
        cout << d << " " << r << endl;
    };  
    static_assert( std::is_same< Type , decltype( callback ) >::value, "Callbacks should be same!" );


    a(15.7f, 15);   
}

因为在lambda的第一个参数是int的情况下-代码将以1个警告进行编译.如何保护代码?

Because in case of first parametr of lambda would be int - code would compile with 1 warning. How to protect code?

推荐答案

callback的类型不是简单的函数.没有捕获的lambda可以衰减为函数指针，但它不是函数指针.这是本地类的实例.

The type of the callback is not a simple function. A lambda with no captures can decay to function pointer but it isn't a function pointer. It's an instance of a local class.

如果要确保lambda的特定函数类型，可以通过将衰减强制为函数指针类型来实现:

If you want to ensure a specific function type for a lambda, you can do that by forcing the decay to function pointer type:

#include <iostream>
#include <functional>
using namespace std;

int main()
{
    using Type = void( float, int );
    std::function<Type> a;
    auto callback = []( float d, int r )
    {
        cout << d << " " << r << endl;
    };  

    // Ensures desired type.
    {
        Type* const func_ptr = callback;   (void) func_ptr;
    }

    a = callback;
    a(15.7f, 15);   
}

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