Hi,
The perldoc says :
If any of the file tests (or either the "stat" or "lstat" operators) are given the special filehandle consisting of a solitary underline, then the stat structure of the previous file test (or stat operator) is used, saving a system call.
So I would expect that the code chunk
Method 1
---------
...
if (-e $path) {
    if (-l $path) {
    print "is a link";
    } elsif (-d _) {
    print "is a direcotry \n";
    } elsif (-f _) {
    print "Is a file";
    } else {
       print "Error";
    }
}

[download]
to be faster than
Method 2
--------- 
if (-e $path) {
    if (-l $path) {
    print "is a link";
    } elsif (-d $path) {
    print "is a direcotry \n";
    } elsif (-f $path) {
    print "Is a file";
     } else {
       print "Error";
    }
}

[download]
as the method1 does one less system call (at least) I expect method1 to be faster.
But when I benchmarked the results tell other wise :
%./reg.pl
Path is file
Path is file
Rate statreuse nostatreuse
statreuse 31847134/s -- -39%
nostatreuse 52083333/s 64% --
%cat reg.pl 
#!/opt/perl_5.8.7/bin/perl
use Benchmark qw(:all);
my $path = "test.lnk";

my $coderef = sub {
    if (-e $path) {
        if (-l $path){
            print "path is link \n";
        } elsif (-d _) {
            print "Path is directory \n";
        } elsif (-f _) {
            print "Path is file \n";
        }
    }
};

my $coderef1 = sub {
    if (-e $path) {
        if (-l $path){
            print "path is link \n";
        } elsif (-d $path) {
            print "Path is directory \n";
        } elsif (-f $path) {
            print "Path is file \n";
        }
    }
};

 my $r = cmpthese( 50000000, {
     statreuse => $coderef->(),
     nostatreuse   => $coderef1->()
 });

[download]
How is that reuse of file stat data slower ?
Do I miss anything here ?
thanks,
sateesh

In reply to Use of undescore to reuse file stat data is slower by Anonymous Monk

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