Question

When I run this code:

int arr[3] = {2,3,4};
char *p;
p = (char*)arr;
printf("%d", *p);
p = p+1;
printf("%d", *p);

The output is 2 and 0. The second result is slightly confusing. Could someone explain why this is happening?

Answer 1

Let’s break this down:

int arr[3] = {2,3,4};

Creates an array of 3 integers. Assuming your system is 32bit little endian this is how it looks in memory:

02 00 00 00 03 00 00 00 04 00 00 00

char *p; p = (char*)arr;

p now points to arr but is a pointer to char*. In other words, p points to the 02.

printf("%d", *p);

You are printing as an int the location referenced by p. So when you dereference p (by writing *p) you are accessing the char (since p is of type char*) referenced by p. Which is 02.

p = p+1;

p now points to the 00 just after 02, because p is char*. So when you add 1, it will move by 1 * sizeof(char) = 1 * 1 = 1 byte in memory.

printf("%d", *p);

You are printing as an int the location referenced by p. So when you dereference p (by writing *p) you are accessing the char (since p is of type char*) referenced by p. Which is 00.

If you wanted to print 3 instead of 0 you have to change your pointer type to int* instead of char*, making the pointer move by 1 * sizeof(int) = 1 * 4 = 4 bytes in memory.