Comment by For_Iconoclasm | Hacker News Reader

Comment by For_Iconoclasm | Hacker News Reader

Well, it got me on the pointer arithmetic questions. I remembered that C automatically handles pointer arithmetic (multiplying by sizeof(type)), however:

Question 2: I didn't think that x+1 would be interpreted as a pointer for some reason, so I guessed 0x7fffdfbf7f01. Wrong.

Question 4: I incorrectly thought that what I remembered about pointer arithmetic would apply here. 1 * sizeof(int) = 0x04, so I guessed 0x7fffdfbf7f04. Wrong.

I don't work in C professionally, but I'd like to not forget things. My error in question 2 shows forgetfulness, and my error in question 4 is from not ever completely mastering every nook & cranny in C.

How did the rest of HN do?

> I incorrectly thought that what I remembered about pointer arithmetic would apply here. 1 * sizeof(int) = 0x04, so I guessed 0x7fffdfbf7f04.

Your pointer arithmetic was correct. What you missed was that &x is a pointer to an array of five ints (so, sizeof(int[5])), not a single int.

I on the other hand didn't realize sizeof(int) is not necessarily 8 on 64-bit machines. You learn something new every day...

On 64-bit machines, int is typically 4 bytes. Longs are typically 8 bytes. But if you really need to assume a certain length, use the typedefs from ctype.h. (For example, if you wanted a 4 byte signed integer, you would say int32_t. An 8 byte unsigned integer would be uint64_t.)

So for question 4 it was a matter of realizing that the type for &x is a pointer to an array of 5 ints:

  int (*y)[5];
  y = &x;
  printf("%p, %p\n", y, y+1);

I got all questions right, although I was not confident (so sizeof(int)is 4).

Reply on news.ycombinator.com