问题描述
[原标题提到sizeof的功能。的]
我尝试了这些，他们都工作：
I tried these and they all worked:
char *p;
printf("Size of *p is %d\n",sizeof(*p)); //result =1
printf("Size of p is %d\n",sizeof( p)); //result =4
printf("Size of p is %d\n",sizeof(&p)); //result =4
我不知道为什么第一个printf是1，第2和第3是4？
所以可以sizeof的实际上可以采取什么样的参数？
I wonder why the first printf is 1, the 2nd and 3rd is 4?So what arguments can sizeof can actually take?
推荐答案
这需要一个类型。
的sizeof（char）的始终为1。变量 P 本身就是一个指针，你的平台上，有一个大小4.然后你做＆安培; P 或一个指针的指针，其中也有一个尺寸为4
sizeof(char) is always one. The variable p itself is a pointer, and on your platform that has a size of 4. Then you do &p, or a pointer to a pointer, which also has a size of 4.
在大多数现代桌面系统，32位架构将有4个字节的指针，而64位架构将有8个字节的指针。
On most modern desktop systems, a 32-bit architecture will have 4 byte pointers, while a 64-bit architecture will have 8 byte pointers.
的sizeof 本身是一个关键字，在编译时解决，而不是一个函数。在C99，数组可以是可变长度，和sizeof会等到运行时来解决这个尺寸。
sizeof itself is a keyword, resolved at compile-time, not a function. In C99, arrays can be variable length, and sizeof will wait until run-time to resolve this size.
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