sizeof operator giving output false while using in this way

Question

I am not able to understand why this piece of code is giving output False:

   if (sizeof(int) > -1)
       printf("True");
   else
       printf("False");

As I tried to print what sizeof(int) is returning is 4.

I'd like to understand in which case this comparison makes sense? Have you ever heard of any type which has negative size? ;-) — Alfe, Jul 16 '14 at 10:23
Don't write such nonsense in the first place, I'd rather say. — Alfe, Jul 16 '14 at 10:26
It's a nice reminder why `x > -1` ≠ `x >= 0` even for whole numbers like integers. — Alfe, Jul 16 '14 at 10:27
look here http://stackoverflow.com/questions/5416414/signed-unsigned-comparisons — Jayesh Bhoi, Jul 16 '14 at 10:30

score 3 · Answer 1 · answered Jul 16 '14 at 10:23

3

The result of the sizeof operator has type size_t. Your -1 is a signed int. When the two are compared, the latter is converted to size_t, which results in a rather large unsigned value.

answered Jul 16 '14 at 10:23

Nisse Engström

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It in fact results in the maximum unsigned value for the type that is `size_t` – Ivaylo Strandjev Jul 16 '14 at 10:26

score 1 · Answer 2 · answered Jul 16 '14 at 10:22

By standard sizeof returns an unsigned integer type size_t. Although the exact type is implementation defined it is certain to be unsigned. When you try to compare it to the signed integer -1, -1 gets converted to max value of this type(try writing (unsigned)-1 and examine the value) and thus the comparison is false.

sizeof operator giving output false while using in this way

2 Answers2