从不兼容的指针类型(指向函数的指针)初始化

Question

我完全陷入了困境.我在3个文件中有以下代码:

文件mixer_oss.c

#include "mixer.h"

static char *devices[] = SOUND_DEVICE_NAMES;

static char **oss_get_device(void)
{
    int i, o, devs, res;
    char **result;

    if ((ioctl(fd, SOUND_MIXER_READ_RECMASK, &devs)) == -1) {
        return NULL;
    } else {
        result = malloc(sizeof(char*)*SOUND_MIXER_NRDEVICES);
        o = 0;
        for (i=0; i < SOUND_MIXER_NRDEVICES; i++) {
                res = (devs >> i)%2;
                if (res) {
                    result[o] = malloc(strlen(devices[i])+1);
                    sprintf(result[o], "%s", devices[i]); 
                    o++;
                }
                result[o] = NULL;   
            }
    }

    return result;
}

struct mixer oss_mixer = {
    .get_device = oss_get_device,
};

文件mixer.h

#ifdef __cplusplus
extern "C" {
#endif

struct mixer
{
    char (* get_device) (void);
};

#pragma weak oss_mixer
extern struct mixer oss_mixer;


#ifdef __cplusplus
};
#endif

file mixer.c

#include "mixer.h"

static char null_get_device(void)
{
}

static struct mixer *mixers[] = {
    &oss_mixer,
    &null_mixer
};

现在,当我编译代码时,这就是我得到的

mixer-oss.c: warning: initialization from incompatible pointer type [enabled by default]
.get_device = oss_get_device,
^

warning: (near initialization for ‘oss_mixer.get_device’) [enabled by default]

请你帮助我好吗？

提前致谢.

Answer 1

char (* get_device) (void);声明一个指向函数的指针,该函数不带参数并返回一个char.

static char **oss_get_device(void)是一个不带参数的函数,并返回指向指针的指针char.

你的函数指针应该像这样声明:

char ** (*get_device)(void);

现在它兼容oss_get_device.