C函数指针混乱 [英] C function pointer confusion

问题描述

这两个声明有什么区别:

What is the difference between these 2 declaration:

int operate(int (*func)(int, int), int a, int b){
    return (*func)(a, b);
}

和

int operate(int func(int, int), int a, int b){
        return func(a, b);
 }

这两个似乎也等效:operate(sum, 1, 1)和operate(&sum, 1, 1)

These two also seems to be equivalent: operate(sum, 1, 1) and operate(&sum, 1, 1)

如果我将函数sum作为2个数字的函数传递给func，则结果仍然相同.为什么?

If I pass function sum as a function of 2 numbers in the place of func, the result are still the same. Why?

推荐答案

§6.7.5.3/8:

§6.7.5.3/8:

将参数声明为函数返回类型"声明调整为指向函数返回类型的指针"，如6.3.2.1.所示.

A declaration of a parameter as ‘‘function returning type’’ shall be adjusted to ‘‘pointer to function returning type’’, as in 6.3.2.1.

换句话说，两个函数声明是相同的.

In other words, the two function declarations are identical.

就函数调用而言，第6.5.2.2/3节:

As far as the function call goes, §6.5.2.2/3:

后缀表达式，后跟括号()，其中可能为空，以逗号分隔 表达式列表是一个函数调用.后缀表达式表示被调用的函数.

A postfix expression followed by parentheses () containing a possibly empty, comma-separated list of expressions is a function call. The postfix expression denotes the called function.

由于func(a, b);和(*func)(a, b)都是后缀表达式，后跟括号，因此它们都是函数调用.由于func和(*func)都指定相同的功能，因此它们都调用相同的功能.

Since both func(a, b); and (*func)(a, b) are postfix expressions followed by parentheses, they're both function calls. Since func and (*func) both designate the same function, they both call the same function.

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