Why does sizeof(*"327") return 1 instead of 8 on a 64 bit system?

Question

 printf("%lu \n", sizeof(*"327")); 

I always thought that size of a pointer was 8 bytes on a 64 bit system but this call keeps returning 1. Can someone provide an explanation?

Answer 1

Putting * before a string literal will dereference the literal (as string literal are array of characters and will decay to pointer to its first element in this context). The statement

printf("%zu \n", sizeof(*"327"));  

is equivalent to

printf("%zu \n", sizeof("327"[0]));   

"327"[0] will give the first element of the string literal "327", which is character '3'. Type of "327", after decay, is of char * and after dereferencing it will give a value of type char and ultimately sizeof(char) is 1.

Answer 2

The statement:

printf("%lu \n", sizeof(*"327")); 

actually prints the size of a char, as using * dereferences the first character of string 327. Change it to:

char* str = "327"; printf("%zu \n", sizeof(str)); 

Note that we need to use %zu here, instead of %lu, because we are printing a size_t value.