函数指针的理解

Posted 2021-07-16 5念since

因为linux内核中一些电源管理相关的代码比较难懂，所以写下了测试代码，共更新迭代了四个版本

1、中规中矩的函数指针的使用

demo中*pfunc与add和sub函数返回值及参数都相同，可以使用其指向这两个函数，当然一般很少能够看到这么直白的使用方法。

#include <stdio.h>

int add(int a, int b);
int sub(int a, int b);
int (*pfunc)(int a, int b);

int add(int a, int b)
{
    return a + b;
}

int sub(int a, int b)
{
    return a - b;
}

int main()
{
    int a=1;
    int b = 2;
    int c;

    pfunc = add;

    c = (*pfunc)(a, b);

    printf("add c = %d\\n", c);

    pfunc = sub;

    c = (*pfunc)(a, b);

    printf("sub c = %d\\n", c);
}

2、使用typedef

#include <stdio.h>

int add(int a, int b);
int sub(int a, int b);
typedef int (*pfunc)(int a, int b);

int add(int a, int b)
{
    return a + b;
}

int sub(int a, int b)
{
    return a - b;
}

pfunc calc_func(char op)
{
    switch(op)
    {
        case '+':
	return add;
	case '-':
	return sub;
	default:
	return NULL;
    }
}

int calc(int a, int b, char op)
{
    pfunc fp = calc_func(op);

    if(fp)
	return fp(a, b);
    else
	printf("can't match.\\n");
}

int main()
{
    int a=1;
    int b = 2;
    int c;

    c = calc(a, b, '+');

    printf("add c = %d\\n", c);


    c = calc(a, b, '-');

    printf("sub c = %d\\n", c);

    c = calc(a, b, '*');
}

3、引入结构体以及变参

#include <stdio.h>

int add(int a, int b);
int sub(int a, int b);
typedef struct calculate{
    int (*sfunc)(int);
    int (*pfunc)(int a, int b);
} g_calc;
typedef int (*pfunc)(int a, int b);

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

int add(int a, int b)
{
    return a + b;
}

int sub(int a, int b)
{
    return a - b;
}

pfunc calc_func(char op)
{
    switch(op)
    {
        case '+':
	return add;
	case '-':
	return sub;
	default:
	return NULL;
    }
}

int calc(int a, int b, char op)
{
    g_calc s_calc = {
        .sfunc = square,
        .pfunc = calc_func(op),
    };
    printf("square a:%d.\\n", s_calc.sfunc(a));
    if(s_calc.pfunc)
	    return s_calc.pfunc(a, b);
    else
	printf("can't match.\\n");
}

int main()
{
    int a=1;
    int b = 2;
    int c;

    c = calc(a, b, '+');

    printf("add c = %d\\n", c);


    c = calc(a, b, '-');

    printf("sub c = %d\\n", c);

    c = calc(a, b, '*');
}

4、最终的使用方法

#include <stdio.h>

int add(int a, int b);
int sub(int a, int b);

typedef struct calculate{
    char s;
    int (*pfunc)(int a, int b);
} g_calc;

typedef int (*pfunc)(int a, int b);

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

int add(int a, int b)
{
    return a + b;
}

int sub(int a, int b)
{
    return a - b;
}

enum calc_func {
    FUNC1_ADD,
    FUNC2_SUB,
    FUNCN_MAX,
};

g_calc s_calc[] = {
    [FUNC1_ADD] = {
        .s = '+',
        .pfunc = add,
    },
    [FUNC2_SUB] = {
        .s = '-',
        .pfunc = sub,
    },
};


int g_calc_func(int a, int b, char op)
{
   int i;
   int match_id = -1;
//在这遇到问题，使用switch case无法匹配
//case label does not reduce to an integer constant
    for(i=0; i < ( sizeof(s_calc) / sizeof(s_calc[0]) ); i++) {
        if( s_calc[i].s == op ) {
            match_id = i;
            break;
        }
    }
   
    if( match_id != -1 )
        return s_calc[match_id].pfunc(a,b);
    else
        printf("can't match.\\n");

    return -1;
}

int main()
{
    int a=1;
    int b = 2;
    int c;

    printf("%d.\\n", g_calc_func(1, 2, '+'));
    printf("%d.\\n", g_calc_func(1, 2, '-'));
    printf("%d.\\n", g_calc_func(1, 2, '*'));
}

参考博客