C - 函数参数前的预期声明说明符或 ‘...’

Question

我的标题定义了以下代码:

typedef uint8_t EnrollT(uint16_t test1, uint16_t test2);
typedef void ChangeT(uint64_t post1, uint8_t post2);

struct ClusterT * ClientAlloc(EnrollT *, ChangeT *);

我已经实现了这两个函数并将它们传递给我的 c 文件中的 ClientAlloc()，如下所示:

ClientAlloc(Enroll, Change);

但是，当我编译源代码时，会弹出错误。

expected declaration specifiers or ‘...’ before ‘enroll’
expected declaration specifiers or ‘...’ before ‘change’

这里有什么我可能会遗漏的吗？

对于 EnrollT 和 ChangeT，我在我的代码中声明:

uint8_t Enroll(uint16_t test1, uint16_t test2){...};
void Change(uint64_t post1, uint8_t post2){...};

对于 ClienAlloc:

struct ClusterT * ClientAlloc(Enroll, Change){... return something};

Answer 1

您正在将 Enroll 和 Change 函数的地址传递给 ClientAlloc

然后是你的

struct ClusterT * ClientAlloc(Enroll, Change){... return something}

必须是

struct ClusterT *ClientAlloc(EnrollT *p, ChangeT *q){... return something}

示例代码是:

#include <stdint.h>
#include <stdlib.h>

typedef uint8_t EnrollT(uint16_t test1, uint16_t test2);
typedef void ChangeT(uint64_t post1, uint8_t post2);

struct ClusterT *ClientAlloc(EnrollT *p, ChangeT *q)
{
   return NULL;
}

uint8_t enroll(uint16_t test1, uint16_t test2)
{
    return 0;
}

void change(uint64_t post1, uint8_t post2)
{

}

int main(void) {

    ClientAlloc(enroll, change);

    return 0;
}