C:如何访问存储在void指针(void *)中的函数指针? [英] C: How to access a function pointer stored in a void pointer (void *)?

问题描述

对于如何访问存储在void指针(void *)中的函数指针感到困惑.

Confused as to how one can access a function pointer stored in a void pointer (void *).

假设您有这个:

void *functions[] =
{
    &sqrt,         // int ft_sqrt(int nb);
    &power,
    &logN,
    &factorial;
};

// An array of void pointers, each storing a function pointer.

---

如果我想访问sqrt函数，我的猜测将是:

---

If I wanted to access the sqrt function, my guess would be the following:

(int (*)(int)) functions[0](x)

但是我的猜测是错误的:

But my guess is wrong:

error: called object type 'void *' is not a function or function pointer

那么如何访问这些功能之一?

So how would one access one of these functions ?

推荐答案

这是 operator优先级 :函数调用操作符的优先级高于强制转换操作符.

It's a matter of operator precedence: The function call operator have higher precedence than the casting operator.

这意味着您的表达式(int (*)(int)) functions[0](x)确实等于(int (*)(int)) (functions[0](x)).

That means your expression (int (*)(int)) functions[0](x) is really equal to (int (*)(int)) (functions[0](x)).

您需要在正确的位置显式添加括号以强制转换指针:((int (*)(int)) functions[0])(x).

You need to explicitly add parentheses in the correct places to cast the pointer: ((int (*)(int)) functions[0])(x).

一个更好的IMO解决方案是使用一个指向函数的指针数组 ，因此该数组元素已经是正确的类型:

A much better solution IMO would be to have an array of pointers to functions, so the array elements already is of the correct type:

typedef int (*function_ptr)(int);

function_ptr functions[] = { ... };

然后不需要强制转换:functions[0](x).

Then no casting is needed: functions[0](x).

然后，您也可以免受伦丁的答案中提到的问题.

Then you also would be safe from the issues mentioned in the answer by Lundin.

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