Using Bitwise Operators to get Bit Values in C

Question

I am writing a program in C where I am comparing two bytes of data, and then seeing if the bytes are different, and if so, at which bits.

This is what I have so far:

int var1 = 81;  //Binary: 0101 0001
int var2 = 193; //Binary: 1100 0001
int diff = var1 ^ var2; //diff = 1001 0000 / 144

Basically I know how to use the XOR bitwise operator to see which bits are different between the two variables, but from here I don't know how to use diff to figure out which bits are the differences. For example, in my above code I'd want to use diff to output "Bit 5 and Bit 8 are different".

They're (usually) not bits 5 and 8, but 4 and 7.

Antti Haapala
– Antti Haapala

2018-02-18 16:39:31 +00:00
Commented Feb 18, 2018 at 16:39 — Antti Haapala
– Antti Haapala, Commented Feb 18, 2018 at 16:39

user2736738 · Accepted Answer · 2018-02-18 16:44:24Z

3

You can use a for loop to get that idea and make bitwise AND with 1 properly left shifted to get the set bits positions

for(size_t i = 0; i < sizeof(int)*8; i++){
  if( diff & (1U << i))
    printf("%zu is different\n",i+1);
}

edited Feb 18, 2018 at 16:44

answered Feb 18, 2018 at 16:36

user2736738

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11 Comments

Pablo Over a year ago

damn, you beat me, I discarded my answer as it was the same :)

Skitzafreak Over a year ago

What is the 1U inside the if statement? I haven't seen that before.

Pablo Over a year ago

@Skitzafreak it means that the 1 should be consider as an unsigned int.

user2736738 Over a year ago

@chux.: Is it wise to use CHAR_BIT..almost all architectures today have CHAR_BIT as 8

chux Over a year ago

That is a style issue. Style issues are best determined by your group/company style guide as it is more important to follow that than spend much effort doing what is "best".

|

chux · Accepted Answer · 2018-02-18 18:44:12Z

2

Far easier to start with unsigned types when doing bit manipulations.

As @coderredoc inquired about solutions across various platforms, even uncommon ones:

Using int:

When int diff is negative, conversion to an unsigned (via masking with an unsigned) may change its bit pattern.

An int may have more than 8 bits per "byte". Diminishes correctness of sizeof(int)*8.

Various integer types may have padding (rare). Diminishes correctness of sizeof(int)*CHAR_BIT.

// OP wants to report first bit index as 1.  0 is more common.
#define BIT_REPORT_OFFSET 0

int bit_position = 0;
int mask;
do {
  mask = 1 << bit_position;
  if (diff & mask) {
    printf("Bit %d\n", bit_position + BIT_REPORT_OFFSET);
  }
  bit_position++;
} while (mask < INT_MAX/2);

if (diff < 0) {
  printf("Bit %d\n", bit_position + BIT_REPORT_OFFSET);
}

For maximum portability, avoid changing types, changing the value of diff and use constants from <limits.h> rather than compute them.

edited Feb 18, 2018 at 18:44

answered Feb 18, 2018 at 18:32

chux

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1 Comment

Skitzafreak Over a year ago

Thank you very much for the well thought out answer. I learned a lot in that.

ensc · Accepted Answer · 2018-02-18 16:42:10Z

0

use unsigned int instead of int; then you can use

for (unsigned int pos = 0; diff; ++pos) {
     if (diff & 1)
         printf("difference in pos %u\n", pos);
     diff >>= 1;
}

or

while (diff) {
    int pos = ffs(diff);
    printf("difference in pos %d\n", pos);

    diff &= ~(1u << pos);
}

answered Feb 18, 2018 at 16:42

ensc

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Comments

Achal · Accepted Answer · 2018-02-18 16:42:46Z

0

To get the different bits position, lets say you have 4 byte integer

for(int bit_index  = sizeof(diff) - 1; bit_index >= 0;bit_index-- ) {
 if((diff >> bit_index & 1) == 1 ){ /* if particular bit is 1, that bit_index value you can use */ 
  printf("[%d] bit is different or 1 \n",bit_index);
}

edited Feb 18, 2018 at 16:42

answered Feb 18, 2018 at 16:37

Achal

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