Re^2: To || or not to or and why.

A good example of the difference in precedence of or and || is in the common perl idiom of

open FILE, "file" or die;
[download]

it is not the same as

open FILE, "file" || die;
[download]

since || has a higher precedence than open the "file" || die gets evaluated before the open. So it will not die even if the open fails. or has a lower precedence than open so the open gets evaluated first in the first example. To have the same effect using || you have to use

open(FILE, "file") || die;
[download]

flounder

Comment on Re^2: To \|\| or not to or and why. Select or Download Code

Replies are listed 'Best First'.
Re: Re^2: To \|\| or not to or and why. by krisahoch (Deacon) on Sep 16, 2003 at 16:09 UTC
Flounder, Thank you for that explaination. So if I am reading it right, then the would following be true? The following functions set the value of the return to 'default' by default. Then if was a parameter passed, $value gets set to that parameter. The differences between the two are: The juxteposition of the shift() statement and 'default' The use of the '\|\|' and 'or' operators `sub setValue_takeOne{ my $value = shift() \|\| 'default'; return $value; } sub setValue_takeTwo{ my $value = 'default' or shift(); return $value; }` [download]	[reply] [d/l]
Re^4: To \|\| or not to or and why. by flounder99 (Friar) on Sep 17, 2003 at 20:54 UTC
No. `my $value = shift() \|\| 'default';` [download] and `my $value = shift() or 'default';` [download] will not work the same. `my $value = shift() \|\| 'default';` [download] will set `$value` to the value returned by `shift()` if it is true and to 'default' if it is false. `my $value = shift() or 'default';` [download] will set `$value` to the value returned by `shift()` no matter what! Here is an example using a handy tool `B::Deparse` that will show you how the compiler read your program, the `-p` adds extra parentheses to make it very clear. `#temp.pl $value = $x or $y or $z; $value = $x \|\| $y \|\| $z; $value = $x or $y \|\| $z; $value = $x \|\| $y or $z;` [download] then run `perl -MO=Deparse,-p temp.pl` [download] it outputs `((($value = $x) or $y) or $z); ($value = (($x \|\| $y) \|\| $z)); (($value = $x) or ($y or $z)); (($value = ($x \|\| $y)) or $z); temp.pl syntax OK` [download] if you notice `$value = $x or $y or $z;` [download] and `$value = $x or $y \|\| $z;` [download] just set `$value` to the value of `$x` no matter what the value of `$y` or `$z` because `or` has a lower precidence than `=`. Did you notice that `$value = $x \|\| $y or $z;` [download] will just set `$value` to the value of `$x \|\| $y` no matter what the value of `$z`? -- flounder	[reply] [d/l] [select]

Replies are listed 'Best First'.
Re: Re^2: To \|\| or not to or and why. by krisahoch (Deacon) on Sep 16, 2003 at 16:09 UTC
Flounder, Thank you for that explaination. So if I am reading it right, then the would following be true? The following functions set the value of the return to 'default' by default. Then if was a parameter passed, $value gets set to that parameter. The differences between the two are: The juxteposition of the shift() statement and 'default' The use of the '\|\|' and 'or' operators `sub setValue_takeOne{ my $value = shift() \|\| 'default'; return $value; } sub setValue_takeTwo{ my $value = 'default' or shift(); return $value; }` [download]	[reply] [d/l]
Re^4: To \|\| or not to or and why. by flounder99 (Friar) on Sep 17, 2003 at 20:54 UTC
No. `my $value = shift() \|\| 'default';` [download] and `my $value = shift() or 'default';` [download] will not work the same. `my $value = shift() \|\| 'default';` [download] will set `$value` to the value returned by `shift()` if it is true and to 'default' if it is false. `my $value = shift() or 'default';` [download] will set `$value` to the value returned by `shift()` no matter what! Here is an example using a handy tool `B::Deparse` that will show you how the compiler read your program, the `-p` adds extra parentheses to make it very clear. `#temp.pl $value = $x or $y or $z; $value = $x \|\| $y \|\| $z; $value = $x or $y \|\| $z; $value = $x \|\| $y or $z;` [download] then run `perl -MO=Deparse,-p temp.pl` [download] it outputs `((($value = $x) or $y) or $z); ($value = (($x \|\| $y) \|\| $z)); (($value = $x) or ($y or $z)); (($value = ($x \|\| $y)) or $z); temp.pl syntax OK` [download] if you notice `$value = $x or $y or $z;` [download] and `$value = $x or $y \|\| $z;` [download] just set `$value` to the value of `$x` no matter what the value of `$y` or `$z` because `or` has a lower precidence than `=`. Did you notice that `$value = $x \|\| $y or $z;` [download] will just set `$value` to the value of `$x \|\| $y` no matter what the value of `$z`? -- flounder	[reply] [d/l] [select]

Flounder,