Re: Removing Trailing Whitespace: Multiple ways.

[perlmonkey]

If it is speed you are after, you can always put on your C hat:

use Benchmark;
use Inline C;

timethese (1000000, {
'regex'  => q{
  my $foo = "test    ";
  $foo =~ s/\s+$//;
},
'inline' => q{  
  my $foo = "test    ";
  rmsp($foo);
}
});
  
__END__
__C__
void rmsp(char * str) {
  int i = strlen(str) - 1; 
  while( i >= 0 && str[i] == ' ' ) {
    str[i--] = '\0';
  }
}

Results:

Benchmark: timing 1000000 iterations of inline, regex...
    inline:  4 wallclock secs ( 4.39 usr +  0.03 sys =  4.42 CPU)
     regex:  8 wallclock secs ( 6.84 usr +  0.03 sys =  6.87 CPU)

[no_slogan]

That would be really cool, but rmsp() leaves the string null-padded instead of removing the extra blanks.

[perlmonkey]

Yah, so I was lazy and I cheated a bit. $foo will look good when you print it though. What can I say, there are reasons I am a perl programmer.

But to be all official like and actually update the scalar value to be a proper string of length 4 you can use this code for rmsp:

void rmsp(SV * sv) {
  char * end;
  int length;
  if( !SvPOK(sv) ) return;
  end = SvEND(sv);
  length = SvCUR(sv);
  end--; /* skip \0 */
  while( *end == ' ' && length >=0 ) { 
    end--;
    length--;
  }
  if( length >= 0 ) SvCUR_set(sv, length);
}

And the new results I got are:

Benchmark: timing 1000000 iterations of inline, regex...
    inline:  6 wallclock secs ( 4.19 usr +  0.01 sys =  4.20 CPU)
     regex:  9 wallclock secs ( 7.71 usr +  0.02 sys =  7.73 CPU)