Sizes of arrays declared with pointers

Question

char c[] = \"Hello\";
char *p = \"Hello\";

printf(\"%i\", sizeof(c)); \\\\Prints 6
printf(\"%i\", sizeof(p)); \\\\Prints 4

My question is:

Why

Answer 1

The two operands to sizeof have different types. One is an array of char, the other a pointer to char.

The C standard says that when the sizeof operator is applied to an array, the result is the total number of bytes in the array. c is an array of six char including the NUL terminator, and the size of char is defined to be 1, so sizeof (c) is 6.

The size of a pointer, however, is implementation-dependent. p is a pointer to char. On your system, the size of pointer to char happens to be 4 bytes. So that's what you see with sizeof (p).

If you try sizeof(*p) and sizeof(*c), however, they will both evaluate to 1, because the dereferenced pointer and the first element of the array are both of type char.