以下程序的输出将为您提供有关类型的大小和类型指针的一些提示和理解。
#include <stdio.h>
int main(void)
{
int p1[10];
int *p2[10];
int (*p3)[10];
printf("sizeof(int) = %d\n", (int)sizeof(int));
printf("sizeof(int *) = %d\n", (int)sizeof(int *));
printf("sizeof(p1) = %d\n", (int)sizeof(p1));
printf("sizeof(p2) = %d\n", (int)sizeof(p2));
printf("sizeof(p3) = %d\n", (int)sizeof(p3));
return 0;
}
int p[10]; => 10 consecutive memory blocks (each can store data of type int) are allocated and named as p
int *p[10]; => 10 consecutive memory blocks (each can store data of type int *) are allocated and named as p
int (*p)[10]; => p is a pointer to an array of 10 consecutive memory blocks (each can store data of type int)
现在回答你的问题:
>> in the first code p points to an array of ints.
>> in the second code p points to an array of pointers.
你是对的。代码中:2、要获取p指向的数组大小,需要传入基地址
printf("%d", (int)sizeof(p));
而不是以下内容
printf("%d", (int)sizeof(*p)); //output -- 4
以下是等效的:
*p, *(p+0), *(0+p), p[0]
>> what's the difference between p[10] and
>> (*p)[10]...they appear same to me...plz explain
以下是您的另一个问题的回答:
int p[10];
_________________________________________
| 0 | 1 | 2 | | 9 |
| (int) | (int) | (int) | ... | (int) |
|_______|_______|_______|_________|_______|
int (*p)[10]
_____________________
| |
| pointer to array of |
| 10 integers |
|_____________________|