What arguments does the sizeof operator take in C?
arguments, c, parameters, pointers, sizeof
Solution
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.
Problem
[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?