C從特定函數指針轉換為不太具體的函數指針

Question

假設我在C中有以下代碼：

void general(void (*function)(void *something), void *something);
void funct_type1(struct something1 *something);
void funct_type2(struct something2 *something);
general(funct_type1, something1);
general(funct_type2, something2);

如何在沒有指針轉換警告的情況下編譯它，而無需手動抑制某些錯誤？

稍后編輯：我正在嘗試不修改funct_type1（）和func_type2（）定義。

Answer 1

正確的解決方法是將特定於實現的東西折疊到抽象接口“后面”的函數中，即：

void funct_type1(void *s1)
{
 struct something1 *s = s1;
 /* ... rest of function ... */
}

否則，您將不得不在對general()的調用中強制轉換函數指針。

Answer 2

你可以使用void *

void general(void (*function)(void *something), void *something);
void funct_type1(void* something);
void funct_type2(void* something);
general(funct_type1, &test1);
general(funct_type2, &test2);

哪里

void funct_type1(void* something)
{
   struct something1 *castedPointer = something;

//...YOUR STUFF
}

或者您可以使用union對結構進行分組

#include <stdio.h>

struct something1
{
    int a;
};

struct something2
{
    int b;
};

union something3
{
    struct something1 s1;
    struct something2 s2;
};

void general(void (*function)(union something3 *something), union something3* something)
{
    if (function != NULL)
        function(something);
}

void funct_type1(union something3* something)
{
    something->s1.a = 1;

    printf("%s - something->s1.a = %d\n", __func__, something->s1.a);
}

void funct_type2(union something3* something)
{
    something->s2.b = 3;

    printf("%s - something->s2.b = %d\n", __func__, something->s2.b);
}

int main()
{
    union something3 test1;
    union something3 test2;

    test1.s1.a = 0;
    test2.s2.b = 0;

    general(funct_type1, &test1);
    general(funct_type2, &test2);

    printf("%s - test1.s1.a = %d\n", __func__, test1.s1.a);
    printf("%s - test2.s2.b = %d\n", __func__, test2.s2.b);

}