Re: reliable get position of leftmost bit in large integers

[hdb]

Should your code not be:

perl -e 'for my $i (1..63) {if(int(log(2**($i-1))/log(2)) != $i-1) {pr
+int "$i\n"}}'

which gives no output for me.

UPDATE: Apologies, I think I get it now, the leading one of 2^i - 1 should be in the same place as the leading one for 2^(i-1) which is what you are testing.

My next thought was to use length(sprintf("%b",$x)) but that does not work either:

for my $i (1..63) {
    if(length(sprintf("%b",2**$i-1)) != length(sprintf("%b",2**($i-1))
+)) {
    print"$i\n";
    printf("%b\n",2**($i-1));
    printf("%b\n",(2**$i)-1);
    }
}

[wollmers]

No, because the problem triggers on vectors filled with on-bits. But the exponentiation ** also seems to have a problem (thanks your remark):

$ perl -e 'print sprintf("%b",2**53-1),"\n"'
11111111111111111111111111111111111111111111111111111
$ perl -e 'print sprintf("%b",2**54-1),"\n"'
1000000000000000000000000000000000000000000000000000000
$ perl -e 'print sprintf("%b",2**(53-1)),"\n"'
10000000000000000000000000000000000000000000000000000

[Eily]

The problem is that ** returns a double (floating point), which explains the lack of precision. You should use 1<<54 like BrowserUk did in his answer.

[BrowserUk]

You should use 1<<54 like BrowserUk did in his answer.

Even old dogs can learn :)