Generating particular bit pattern using bitwise operators

Question

Here I want to generate a bit pattern to set n digits equal to 1 starting from position p. Digits are numbered from 0 to 31. Following is what I did.

int bitPattern(int n, int p) {
    int hex, num1, num2;
    hex = 0x80000000;
    num1 = (hex >> (31 - p));
    num2 = (hex >> (31 - (n+p)));
    return num1 ^ num2;
}

Example:

bitPattern(6, 2) should return 
..000011111100

Any alternate solutions with less operators ?

You're invoking undefined behaviour when n+p > 31. If that's guaranteed to never be the case, you still have implementation-defined behaviour right-shifting a negative number. But if you're not afraid of UB, what about ((1 << n) - 1) << p? — Daniel Fischer
– Daniel Fischer, Commented Jun 1, 2013 at 2:11
@DanielFischer in my problem (n+p) won't go beyond 31. and it is easy to identify too... — noufal
– noufal, Commented Jun 1, 2013 at 2:15
Anyway, if you only care about the bits, I'd recommend using an unsigned type, and then ((1 << n) - 1) << p is safe unless n or p are negative or greater than or equal to the width of the type. — Daniel Fischer
– Daniel Fischer, Commented Jun 1, 2013 at 2:18

Sergey Kalinichenko · Accepted Answer · 2013-06-01 02:13:07Z

5

You can do it like this:

return ((1<<n)-1)<<p;

To make n ones at position zero, compute (2^n)-1; recall that 2^n is 1<<n, so the expression becomes ((1<<n)-1). Now you need to add p zeros at the back, so shift the result left by p.

answered Jun 1, 2013 at 2:13

Sergey Kalinichenko

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