I have numbers;

A == 0x20000000
B == 18
C == (B/10)
D == 0x20000004 == (A + C)

A and D are in hex, but I'm not sure what the assumed numeric bases of the others are (although I'd assume base 10 since they don't explicitly state a base.

It may or may not be relevant but I'm dealing with memory addresses, A and D are pointers.

The part I'm failing to understand is how 18/10 gives me 0x4.

Edit: Code for clarity:

*address1 (pointer is to address: 0x20000000)

printf("Test1: %p\n", address1);
printf("Test2: %p\n", address1+(18/10));
printf("Test3: %p\n", address1+(21/10));


Test1: 0x20000000
Test2: 0x20000004
Test3: 0x20000008
  • 1
    What language? Or should we be guessing? – Oded Sep 2 '12 at 19:48
  • I wasn't aware the language was relevant, but if so, the language I'm working with is C. – Hamid Sep 2 '12 at 19:49
  • 3
    It certainly is. Different languages have different literal semantics. – Oded Sep 2 '12 at 19:50
  • I see. Thanks for the head up. In that case, you should take little of what I have in the code blocks above as being literal. It's an arbitrary representation I constructed for the question. – Hamid Sep 2 '12 at 19:52
  • 2
    What is the data type of address1? If address1 is a pointer which points to something whose sizeof happens to be 4 bytes on your platform, then address1 + 1 will calculate an address value offset from address1 by 4 bytes. (Note that int is often 4 bytes in size on a 32-bit desktop platform) – Ben Cottrell Sep 2 '12 at 20:07

Notice some facts:

1) when you add a value to the address it gets increased by that value multiplied by the number of bytes contained in a word, not by simply that value;

2) 18/10 == 1 when it comes to integers;

3) 21/10 == 2 when it comes to integers;

4) word size is 4 in this case (as you notice by the pointer's size, being 32 bit);


0x20000000 + 4 * (18/10) = 0x20000000 + 4 * 1 = 0x20000004
0x20000000 + 4 * (21/10) = 0x20000000 + 4 * 2 = 0x20000008

As Vatine pointed out, it's important understanding that the pointer is incremented by a value multiplied by 4 (i.e a word's size in a 32-bit system) because that's the size of the data type the pointer variable was created for (an int).

  • When you add a value to a pointer, it is incremented by the number of chars necessary to store the type pointed to. This could be 1,2,4,... chars (or, indeed, even 3 chars, but that is rather unlikely on modern machine). – Vatine Sep 3 '12 at 12:00
  • @Vatine Indeed, I should have mentioned that the 4 is a consequence of the pointer being a pointer to int; instead I preferred leaving the answer as concise as possible, but you're right. – Nadir Sampaoli Sep 3 '12 at 22:36
  • 1
    @Vatine: Adding an integer N to a pointer gives you a new pointer value advanced N elements past the location to which the original pointer points. Saying that adding N to a pointer actually adds N * sizeof (whatever) to the pointer implies that pointers are really just integers; they're not. – Keith Thompson Sep 4 '12 at 2:57
  • @KeithThompson: You are correct. I made a short, maybe not as clear as it could have been, clarification to an accepted answer talking about "addresses". – Vatine Sep 4 '12 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.