Question

如果我定义这样的类：

struct _HNum
{
   char *a;
};
typedef struct _HNum HNum;

然后写一个函数：

/*
*Allocates a new HNum with the same value as hnum. It is the caller's
* responsibility to free the returned HNum.
*
* RETURN VALUE:
*   Returns a pointer to the new number, or NULL if the allocation failed.
*/
HNum *HNum_clone(const HNum *hnum)
{
    HNum newNum;
    if(!newNum.a)
    {
        return NULL;
    }
    strcpy(newNum.a, hnum.a);
}

编译器在“hnum”上给出“表达式必须具有类类型”错误在strcpy(newNum.a, hnum.a);行。这有什么问题？

Answer 1

这是因为hnum是一个指针。因此，您需要取消引用它 - 明确地使用星号，或使用->运算符而不是点.：

strcpy(newNum.a, hnum->a);

注意：您需要添加内存分配和return语句，否则您的程序会有未定义的行为：

HNum *HNum_clone(const HNum *hnum)
{
    if(!hnum)
    {
        return NULL;
    }
    HNum *newNum = malloc(sizeof(HNum));
    if (hnum->a) {
        newNum->a = malloc(strlen(hnum->a)+1);
        strcpy(newNum->a, hnum->a);
    }
    else
    {
        newNum->a = NULL;
    }
    return newNum;
}

Answer 2

这段代码好吗？

typedef struct _HNum 
{
    char *a;
}HNum;


/*
 * Allocates a new HNum and sets its value to 0.
 *
 * RETURN VALUE:
 *   Returns a pointer to the new number, or NULL if the allocation failed.
 */
HNum *HNum_alloc()
{
    HNum *newNum = (HNum*)malloc(sizeof(HNum));
    newNum->a = (char*)malloc(2);
    if(!newNum->a) return NULL;
    newNum->a = "0";
    return newNum;
}


/*
 *Allocates a new HNum with the same value as hnum. It is the caller's
 * responsibility to free the returned HNum.
 *
 * RETURN VALUE:
 *   Returns a pointer to the new number, or NULL if the allocation failed.
 */
HNum *HNum_clone(const HNum *hnum)
{
    if(!hnum)
    {
        return NULL;
    }
    HNum *newNum = HNum_alloc();
    if (hnum->a) {
        strcpy(newNum->a, hnum->a);
    }
    else
    {
        newNum->a = NULL;
    }
    return newNum;
}

＆＃34;表达式必须具有类型类型＆＃34;定义的类的错误

2 个答案: