Question

As of what I know about ‘&’ operator, it returns the base address of the operand in memory.

Let us imagine the following scenario (as on my machine):

• sizeof(int) = 4 bytes
• sizeof(float) = 4 bytes
• sizeof(char) = 1 byte

Now, if I write something like this:

void main() {
 int i = 5411;
 int *ip = &i;
 char *c = &i;

 printf("%d",*ip);
 printf("%c",*c);
}

The first printf() should give me 5411. Talking about the second printf(), the base address of i contains 10101001 (higher order 8 bits = 1 byte for char type pointer).
Hence *c should give me 169, which when converted to %c is an invalid character.

But the compiler is giving me ‘#’ or some other valid output. Why is it so ? Any inputs ?

EDIT (taken from the author’s comment on one of the answers):

That was just a dummy case, since I was away from the actual machine.

The actual case is i = 5411

Answer 1

You seems to have trouble understanding how integers are stored in memory. Take 5411 as example.

5411 = 1010100100011

this number 13 binary digits has however, since an int is 32-bit, it must be pad to 32 digits

5411 = 00000000 00000000 00010101 00100011

On a little endian machine (x86, ARM by default), the least significant bytes are stored in the front, so in the memory:

00100011   00010101    00000000    00000000
^
c          c + 1       c + 2       c + 3
ip

Therefore, *c should return 00100011 i.e. 35 ('#').