CODEWITHSUNDEEP
cbit-manipulationbitbitwise-operators
Geek
Geek

Reputation: 27203

How is this bitwise AND operator masking the lower seven order bits of the number?

I am reading The C Programming Language by Brian Kernigan and Dennis Ritchie. Here is what it says about the bitwise AND operator:

The bitwise AND operator & is often used to mask off some set of bits, for example,

    n = n & 0177

sets to zero all but the low order 7 bits of n.

I don't quite see how it is masking the lower seven order bits of n. Please can somebody clarify?

Upvotes: 32

Views: 9367

Answers (5)

Miss J.
Miss J.

Reputation: 391

0177 is an octal value. In octal, each digit is represented by 3 bits from the value 000 to 111. So, 0177 translates to 001111111 (i.e. 001|111|111). This number in 32 bit binary (can be 64 bit too except you need to sign-extend, which will be all 0s in this case) is 00000000000000000000000001111111
(00|000|000|000|000|000|000|000|001|111|111).

Performing a bitwise AND (symbolically, &) with it for a given number will output the lower 7 bits of the number, turning of rest of the digits in the n-bit number to 0.

(since x&0 =0 & x&1=x e.g 0&0=0 ,1&0=0, 1&1=1 0&1=1)

Upvotes: 0

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726809

The number 0177 is an octal number representing the binary pattern below:

0000000001111111

When you AND it using the bitwise operation &, the result keeps the bits of the original only in the bits that are set to 1 in the "mask"; all other bits become zero. This is because "AND" follows this rule:

X & 0 -> 0 for any value of X
X & 1 -> X for any value of X

For example, if you AND 0177 and 0545454, you get

0000000001111111 -- 0000177
0101010101010101 -- 0545454
----------------    -------
0000000001010101 -- 0000154

Upvotes: 37

Levent Divilioglu
Levent Divilioglu

Reputation: 11612

It is already explained that the first '0' used for octal representation of a number in ANSI C. Actually, the number 0177 (octal) is same with 127 (in decimal), which is 128-1 and also can be represented as 2^7-1, and 2^n-1 in binary representation means take n 1's and put all the 1's to the right.

0177 = 127 = 128-1

which is a bitmask;

0000000000000000000000001111111

You can check the code down below;

Demo

#include <stdio.h>

int main()
{
    int n = 0177;   // octal representation of 127
    printf("Decimal:[%d] : Octal:[%o]\n", n, n, n);

    n = 127;        // decimal representation of 127
    printf("Decimal:[%d] : Octal:[%o]\n", n, n, n);

    return 0;
}

Output

Decimal:[127] : Octal:[177]
Decimal:[127] : Octal:[177]

Upvotes: 4

Shafik Yaghmour
Shafik Yaghmour

Reputation: 158529

Since 0177 is an octal literal and each octal number is 3 three bits you have, the following binary equivalents:

7  = 111
1  = 001

Which means 0177 is 001111111 in binary.

Upvotes: 5

ouah
ouah

Reputation: 145899

In C an integer literal prefixed with 0 is an octal number so 0177 is an octal number.

Each octal digit (of value 0 to 7) is represented with 3 bits and 7 is the greatest value for each digit. So a value of 7 in octal means 3 bits set.

Upvotes: 6

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