lowdisc_bitlo0 — Returns the position of the low zero bit base 2 in an integer.
bit = lowdisc_bitlo0 ( n )
the integer to be measured, should be nonnegative.
the position of the low 0 bit, with 1<= bit <= d+1, where d is the number of digits to represent n in base 2.
Consider the number 11 = "1 0 1 1" in binary. The low zero bit is the first zero starting from the right. It is located at the index bit=3. If a number is made only of d ones in base 2, then bit = d+1. This routine is not vectorized, i.e. it does not take a column matrix n as input argument, but only a 1 x 1 matrix. TODO : vectorize this, if possible
// n = 11 is equal to "1 0 1 1" in binary
d = lowdisc_bary ( 11 , 2 )'
// Hence, the low 0 bit is 3
// i.e., from the right, the first zero is at index 3.
bit = lowdisc_bitlo0 ( 11 )
// Compute the low 0 bit for several integers
mprintf("%5s %25s %5s\n","N","Binary","bit");
mprintf("-------------------------------------\n");
nmat = [0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1023 1024 1025];
for n = nmat
bit = lowdisc_bitlo0 ( n );
d = lowdisc_bary ( n , 2 );
mprintf("%5d %25s %5d\n",n,strcat(string(d)," "),bit);
end
Author:
2008-2009 - INRIA - Michael Baudin (Scilab version)
2010 - Digiteo - Michael Baudin