什么是用于的函数类型？ [英] What is a function type used for?

问题描述

给定以下两个 typedef s：

typedef void (*pftype)(int);

typedef void ftype(int);

我理解第一个定义 pftype 一个指向一个具有 int 参数并且不返回任何内容的函数的指针，第二个函数定义 ftype 需要一个 int 参数并不返回任何内容。

I understand that the first defines pftype as a pointer to a function that takes one int parameter and returns nothing, and the second defines ftype as a function type that takes one int parameter and returns nothing. I do not, however, understand what the second might be used for.

我可以创建一个与这些类型匹配的函数：

I can create a function that matches these types:

void thefunc(int arg)
{
    cout << "called with " << arg << endl;
}

然后我可以使用以下函数创建指向这个函数的指针：

and then I can create pointers to this function using each:

int main(int argc, char* argv[])
{
    pftype pointer_one = thefunc;
    ftype *pointer_two = thefunc;

    pointer_one(1);
    pointer_two(2);
}

当使用函数类型时，我必须指定我创建指针。使用函数指针类型，我不。两者都可以作为参数类型互换使用：

When using the function type, I have to specify that I'm creating a pointer. Using the function pointer type, I do not. Either can be used interchangeably as a parameter type:

void run_a_thing_1(ftype pf)
{
    pf(11);
}

void run_a_thing_2(pftype pf)
{
    pf(12);
}

因此，什么是函数类型？

What use, therefore, is the function type? Doesn't the function pointer type cover the cases, and do it more conveniently?

推荐答案

除了你指出的用法之外， （指针的基本类型或对函数的引用），函数类型的最常见用法是在函数声明中：

As well as the use you point out (the underlying type of a pointer or reference to a function), the most common uses for function types are in function declarations:

void f(); // declares a function f, of type void()

code> typedef ：

for which one might want to use a typedef:

typedef void ft(some, complicated, signature);
ft f;
ft g;

// Although the typedef can't be used for definitions:
void f(some, complicated, signature) {...}

并作为模板参数：

std::function<void()> fn = f;  // uses a function type to specify the signature

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