为什么func与C中的&func相同? [英] Why does it seem that func is the same as &func in C?

问题描述

根据

According to the GNU C manual, functions can be called using function pointers like so:

func (j);  /* (*func) (j); would be equivalent. */

所以我在这里的理由是:func本身是func(int)函数的指针.调用func(j)时，您将隐式访问指针func的值(您将移至func所在的存储位置)，其方式与使用具有整数指针的方式相同. ，然后使用*访问存储在内存中该位置的值.这与您可以使用(*func)(j)调用同一函数的事实是一致的.

So my reasoning here is: func itself is a pointer to the func(int) function. When you call func(j), you are implicitly accessing the value of the pointer func (you are moving to the memory location where func is), in the same way as when you have a pointer to an integer, for example, and you access the value stored in that position of the memory using *. That would be consistent with the fact that you can call that same function using (*func)(j).

实际上，在 cprogramming.com 中，他们说您可以拥有指向函数指针的指针.因此，我因此猜测它们的工作原理与任何其他类型的指针一样.

In fact, in cprogramming.com, they say you can have a pointer to a pointer of a function. So I am therefore guessing they work like any other kind of pointers.

但是如果是这样，为什么这段代码可以正常工作?

But if that's the case, why is it that this code works?

#include <stdlib.h>
#include <stdio.h>

void a(int n) {
    printf("%d\n", num);
}

int main() {
    int x = 5;
    void (*func)(int); // Declare a pointer to a function
    func = &a; // Pointer to a pointer to a function
    (*func)(x); // Calls the function (why?)
    func = a; // Pointer to a function
    (*func)(x); // Calls the function (makes sense)
}

此外，如果您致电:

printf("%s\n", (&a == a) ? "True" : "False");

它将打印True！

It prints True!

例如，我确定&foo与&&foo不相同，因此为什么 func与&func相同的原因为何 ?

I am sure that &foo is not the same as &&foo, for example, so why does it seem to be the case that func is the same as &func?

推荐答案

N1570 6.3.2.1 Lvalues, arrays, and function designators says:

4函数指示符是具有函数类型的表达式.除了 当它是sizeof运算符的操作数时，_Alignof 65)或一元&操作符，类型的功能指示符 函数返回类型"被转换为具有 类型为函数返回类型的指针".

4 A function designator is an expression that has function type. Except when it is the operand of the sizeof operator, the _Alignof operator,65) or the unary & operator, a function designator with type ‘‘function returning type’’ is converted to an expression that has type ‘‘pointer to function returning type’’.

此处a和*func是功能指示符，因为它们具有功能类型. &a中的a不会转换为指针，因为它是一元&运算符的操作数，并且该函数的指针由&运算符检索. 另一方面，根据该规则，func = a;中的a被转换为指向该函数的指针. 因此，此代码中的a和&a是等效的.

Here a and *func are function designators because they have function type. a in &a is not converted to a pointer because it is the operand of the unary & operator and the pointer to the function is retrieved by the & operator. On the other hand, a in func = a; is converted to the pointer to the function according to this rule. Therefore a and &a in this code are equivalent.

func(x);中的func也会根据此规则转换为指针.

Also func in func(x); is converted to the pointer according to this rule.

(*func)(x);是:

1. func根据此规则转换为指针
2. *func
中的*取消了指针的引用
3. *func根据此规则转换为指针

1. func is converted to the pointer according to this rule
2. The pointer is dereferenced by * in *func
3. *func is converted to the pointer according to this rule

因此(*func)(x);等同于func(x);.

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