Reputation: 970
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 ?
Upvotes: 5
Views: 1324
Reputation: 726509
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
.
Upvotes: 5