Single questions about file processing in perl

[dima]

Hello all!

The following code is correct :

open(FD,"< my.txt") or die("Error");
while(<FD>)
{
print "$_\n";
exit(0) if ($_eq "STOP");
}

But this code work incorrect :

open(FD,"< my.txt") or die("Error");
$_="NOT_STOP";
while(<FD> and ($_ne "STOP") )
{
print "$_\n";
}
print "$_\n";
exit(0);
Both example are equivalent? Or I wrong?

Furthermore, why this code is incorrect :
while(<FD> and 1)
{
print "$_\n";
}
Thanks!

 

[Paul Lalli]

Hello all!

The following code is correct :

open(FD,"< my.txt") or die("Error");
while(<FD>)
{
print "$_\n";
exit(0) if ($_eq "STOP");
}

But this code work incorrect :

This is a poor error description. You should always tell us HOW it is
incorrect. What did it not do that you expected it to do?
Fortunately, in this case, the answer is plain.

open(FD,"< my.txt") or die("Error");
$_="NOT_STOP";
while(<FD> and ($_ne "STOP") )
{
print "$_\n";
}
print "$_\n";
exit(0);
Both example are equivalent? Or I wrong?

Unfortunately, they are not equivalent. The construct of
while (<FD>){ ... }
is "magical". When <> is the only operation or statement inside the
while() condition, then, AND ONLY THEN, does $_ get assigned the return
value of the <> operation. In ANY other circumstance, you need to
explicitly assign $_.

Furthermore, why this code is incorrect :
while(<FD> and 1)
{
print "$_\n";
}

Same reason as above. The auto-assigning of $_ = <FD> occurs if and
only if <FD> is the ONLY thing inside the while() condition. putting
the additional code ("and 1") in the parentheses prevents this magic.

Hope this helps,
Paul Lalli

 

[Joe Smith]

The following code is correct :

open(FD,"< my.txt") or die("Error");
while(<FD>)
{
print "$_\n";
exit(0) if ($_eq "STOP");
}

No, that is not correct. Since you neglected to chomp(), it prints
out the contents of the file double spaced and does not stop at a
line with "STOP" (unless you have a premature end-of-file).
-Joe