It actually returns an integer with one bit set: the bit to the right of the MSSB. I overlooked this modification when I posted the code. Another modification is that it will not return less than 2. I may have to edit in an unmodified version to fix the original post. The modification is part of the aforementioned binary search variant I'm experimenting with, and I meant to leave that out.

Given the leftmost / most significant bit is #0, the original code is equivalent to: 
my $z=17;

my $m= $z<4? 2:  1<<( 63-( clz($z) +1) );  # 64-bit system integer
my $m= $z<4? 2:  1<<( 31-( clz($z) +1) );  # 32-bit system integer

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EDIT 2024/3/27
I know the original point is moot, but there is always the possibility that a future project will benefit from this idea somehow, and writing these binary chop statements by hand is precarious. The below code generator makes short work of it. Any power-of-2 from 2 to 64 will work as long as your system supports integers that size. 128 and 256 should work as well. Higher powers-of-2 will need a larger bitvector than the one I assigned to $BinFlow.



EDIT 2024/3/30: more comprehensible variable names and comments 
use strict;
use warnings;

####   BINARY CHOP GENERATOR
####   This script generates a single Perl statement
####   in the form of a tree of inline conditionals
####   to find the count of leading zeros in an integer $z
####   with bit width P2 in log(P2)/log(2) steps.
####
####   Disclaimer: for a better solution, use GCC::Builtins qw(clz).


#configuration
use constant Tab         => ' 'x4; # tab width
use constant P2          => 64;    # bit width
                                   # (2, 4, 8, 16, 32, 64, 128, 256)


use constant nSteps      => (log(P2)/log(2) );
use constant MaxInt      => (1<< P2)-1 ;
use constant MASK_0      =>  1<<(P2 -1);
use constant CASE_0_LS   =>     P2 -1;

my $BinFlow=    pack('C9', (0)x9); # bitvector of iterators carries
                           # the binary composition forward
                           # as the loop scales the tree,
                           # printing it line by line.
my $B=1;
my $ls_=0;           # supporting conditional's running left-shift
my $ls=$ls_+(P2>>1); #    "this" conditional's running left-shift


print('my $m=',sprintf('0x%X', MaxInt&( -(1<<($ls-1)) )),'&$z?  ');


while($B>0){
   my $if_else=vec($BinFlow,  $B, 2)++;   #advance conditional $B

   if($if_else==0){                       #BEGIN NEW CONDITIONAL
                                          #(the "if" part)

      if($B<nSteps){                      #not the final step yet?
         $ls_=vec( $BinFlow,++$B, 8)=$ls; #set left-shift as of $B
         $ls+=           P2 >>$B;       

         print("\r\n",   Tab x$B);
         printf('0x%X&$z?  ',  MaxInt&( -(1<<($ls-1)) ) );
      }else{
         # To delineate all 65 unique conditional flow paths, 
         # the 0th and 1st paths require an additional step.
         # These are treated as the most unlikely cases.

         printf("\r\n".(Tab x($B+1)).
               '0x%X&$z?    0'.
               "\r\n".( Tab  x$B). ':'.
               (' 'x(8+((log(MASK_0)/log(16))>>0))), MASK_0 )

                                             if $ls==CASE_0_LS;


         print(1+P2-(P2>>$B)-$ls, ':    ', 1+P2-$ls, "\n");

         vec(      $BinFlow,  $B, 2)=0;   #reset step $B.
         $ls=$ls_;                        #revert left-shift and
         $ls_=vec( $BinFlow,--$B, 8);     #fall back to previous step.
         }
      
   }elsif($if_else==1){                   #RESUME CONDITIONAL
                                          #(the "else" part)
         print(          Tab x$B);
         $ls_=$ls; ++$B;                  #climb back up a step
         $ls-=           P2 >>$B;         #negate shift component B

         printf(':   0x%X&$z?  ', MaxInt&( -(1<<($ls-1)) ) );

   }else{                                 #END CONDITIONAL

         vec(      $BinFlow,  $B, 2)=0;   #reset step $B.
         $ls=$ls_;                        #revert left-shift and
         $ls_=vec( $BinFlow,--$B, 8);     #fall back to previous step.
   }     }

print(";\r\n");

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In reply to Re^2: Most Significant Set Bit by coldr3ality
in thread Most Significant Set Bit by coldr3ality

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