CODEWITHSUNDEEP
cbit-manipulation
zell
zell

Reputation: 10204

Why (3 & 0x1111) = 1?

I do not see why 3 & 0x1111 = 1 ? It seems that:

for any unsigned 32-bit integer i, i & 0x1111 should be i, right?

However when I tried this on ubuntu 14.04, I got 3 & 0x1111=1. Why? 

int main() {
  unsigned int a =3;
  printf("size of a= %lu\n",sizeof(a));
  printf("value of 3 & 0x1111= %d\n",a & 0x1111);

  return 0;
}

Upvotes: 3

Views: 871

Answers (6)

cerkiewny
cerkiewny

Reputation: 2851

The & operator applies the binary and. The 0x means hexadecimal not binary so if we write 0x1111 into a binary we will get: 0001 0001 0001 0001 binary. 3 binary is 011

and 

0001 0001 0001 0001 & 
0000 0000 0000 0011 =
0000 0000 0000 0001 = 1

Upvotes: 1

Sufian Latif
Sufian Latif

Reputation: 13356

In C, any numeric literal starting with 0x is a hexadecimal number. So the bitmask you are using is 1111 in hexadecimal. In the mask, bits #0, #4, #8 and #12 are 1s, and the rest are 0s. That's why you're getting 1.

0x1111 = 0000 0000 0000 0000 0001 0001 0001 0001 in binary
     3 = 0000 0000 0000 0000 0000 0000 0000 0011 in binary
------------------------------------------------
     1 = 0000 0000 0000 0000 0000 0000 0000 0001 after doing biwise AND

If you want to construct a mask with all 1s, in hex, it should be

0xffffffff = 1111 1111 1111 1111 1111 1111 1111 1111

Upvotes: 7

Rowland Shaw
Rowland Shaw

Reputation: 38130

0x1111 is 4369, or as binary: 0001000100010001 So, 3 (0011) masked against that is going to be 0001. Similarly, 19 (0001011) would be 17 (00010001)

Upvotes: 2

dbush
dbush

Reputation: 224362

3d = 3h = 11b

1111h = 0001000100010001b

so:

  0001000100010001b
&               11b
-------------------
                  1b

Upvotes: 4

Eugene Sh.
Eugene Sh.

Reputation: 18371

0x1111 is 0001000100010001 in binary. So 0x1111 & 3 is 0001000100010001 & 0000000000000011 = 0000000000000001

Upvotes: 3

Code Different
Code Different

Reputation: 93181

Convert both of them to binary:

0x1111 = 0001 0001 0001 0001
3      = 0000 0000 0000 0011

When you & them bit by bit, what else do you expect?

Upvotes: 17

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