Question

[Original title referred to ‘sizeof function’.]

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

I wonder why the first printf is 1, the 2nd and 3rd is 4?
So what arguments can sizeof can actually take?

Answer 1

It takes a type.

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.

On most modern desktop systems, a 32-bit architecture will have 4 byte pointers, while a 64-bit architecture will have 8 byte pointers.

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.