C:传递函数指针两次

Question

我想编写一个灵活的 C 程序，允许我执行多种不同的操作。下面是我的代码，但它不起作用。

int result;

Void operation(void (*funcpt1)(int (*funcpt2)(int i)))
{
    (*funcpt1)(funcpt2());
}

Void sum(void (*funcpt)(void))
{
    int i;
    result = 0;
    for(i = 1; i < 10; i++)
    {
        result += (*funcpt)(i);
    }
}

Void multiply(void (*funcpt)(void))
{
    int i;
    result = 1;
    for(i = 1; i < 10; i++)
    {
        result *= (*funcpt)(i);
    }
}

int square(int i)
{
    return i*i;
}

int cube(int i)
{
    return i*i*i;
}

void main(void)
{
    operation(sum(square()));
    printf("%d\n",result);
    operation(sum(cube()));
    printf("%d\n",result);
    operation(multiply(square()));
    printf("%d\n",result);
    operation(multiply(cube()));
    printf("%d\n",result);
}

Answer 1

这是您要找的吗？

#include <stdio.h>

typedef int (*unary_operation)(int);
typedef int (*binary_operation)(int, int);

int add(int lhs, int rhs) {
    return lhs + rhs;
}

int multiply(int lhs, int rhs) {
    return lhs * rhs;
}

int square(int i) {
    return i*i;
}

int cube(int i) {
    return i*i*i;
}

int main() {
    unary_operation unary_op = &cube;
    printf("%d\n", unary_op(3));

    binary_operation binary_op = &add;
    printf("%d\n", binary_op(1, 2));
}

输出为:

27
3