Ç - 前函数参数预期的声明符或“...”

Question

我的头定义了以下代码：

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功能，并在我的C文件传递那些ClientAlloc（）作为如下所示：

ClientAlloc(Enroll, Change); 

但是，当我编译源时，弹出错误。

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

有什么我可能会在这里错过吗？

对于EnrollT和ChangeT，我宣布它在我的代码：

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

要传递到你的Enroll和Change功能ClientAlloc函数地址

然后你的

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

必须

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

一个例子代码：

#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; 
} 

Answer 2

这这里编译罚款：

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

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


struct ClusterT * ClientAlloc(EnrollT *x, ChangeT *y) 
{ 
    (*x)(22,33); 
    return NULL; 
} 


unsigned char enrollfunc(uint16_t test1, uint16_t test2) 
{ 
    return 123; 
} 

void main() 
{ 
    EnrollT *x = enrollfunc; 
    ChangeT *y = NULL; 


    ClientAlloc(x, y); 
}