Hi monks,

In an algorithm I need to test, if the positions of leftmost bits of two bitvectors are equal:

if ( int(log($v1)/log(2))  == int(log($v2)/log(2)) ) { #match }

[download]

But this is unreliable because of the float-precision :

$ perl -e 'for my $i (1..63) {if(int(log(2**$i-1)/log(2)) != $i-1) {pr
+int "$i\n"}}'
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63

[download]

Is there a reliable solution with comparable speed? A split into two 32-bit vectors would be an ugly workaround. Or should I redesign this algorithm LCS_64i() at a higher level?

TIA

Helmut Wollmersdorfer

In reply to reliable get position of leftmost bit in large integers by wollmers

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