[doc] readdir , useless example, not working with "0" directory

[znik3040]

Note: bad example in perl core documentation at https://perldoc.perl.org/functions/readdir

In currently available version, bad line:

while (readdir $dh) {

Description
It won't work properly in any situation. For perl, false is:

1. '' - empty string
2. undef
3. 0 - zero value
4. "0" - text with zero char
5. empty table in scalar context
6. empty hash table inscalar context

Except above, value is interpreted as true.

Example code for verification true/false, no one should be printed:

use 5.010;
say "undef" if undef;
say "empty" if '';
say "0" if 0;
say '"0"' if "0";
my @sometable=();
my %somehash;
say "empty table" if @sometable;
say "empty hash" if %somehash;

when readdir function reads directory or file name "0", in boolean context it is false. finally. after reading "0" by readdir, it returns false to while loop.
But readdir after the last element, returns undef. Then full working version should be corrected to:

while (defined readdir $dh) {

please modify documentation, because this is trap for unwary programmers.
Note, I usually use pragmas warning and strict, and do not use default assigment to $_ variable. Then that line i modify into version:

while (defined my $item=readdir $dh) {

and working with named variable.

[mauke]

For perl, false is:

1. '' - empty string
2. undef
3. 0 - zero value
4. "0" - text with zero char

These are all correct.

5. empty table in scalar context
6. empty hash table inscalar context

Point 5 is just point 3 again: An array in scalar context yields the number of elements, which for an empty array is 0.
Point 6 is just point 3 (or 1) again: A hash in scalar context yields the number of keys, which for an empty hash is 0. (In old perls before v5.25, an empty hash would yield an empty string instead, i.e. point 1.)

when readdir function reads directory or file name "0", in boolean context it is false. finally. after reading "0" by readdir, it returns false to while loop.

This is incorrect. Quoting perldoc -f readdir:

As of Perl 5.12 you can use a bare readdir in a while loop, which will set $_ on every iteration. If either a readdir expression or an explicit assignment of a readdir expression to a scalar is used as a while/for condition, then the condition actually tests for definedness of the expression's value, not for its regular truth value.

This means there are two rewrite rules in place:

1. If the loop condition is just a call to readdir, an implicit assignment to $_ is added: while (readdir ...) is rewritten to while ($_ = readdir ...).
2. If the loop condition is an assignment expression where the left-hand side is a scalar and the right-hand side is a call to readdir, it is implicitly wrapped in defined(): while ($foo = readdir ...) is rewritten to while (defined($foo = readdir ...)).

In other words, the example code is correct as written.

Then that line i modify into version:

while (defined my $item=readdir $dh) {

and working with named variable.

This, on the other hand, is a syntax error.

I'm open to improving the wording of perlfunc/readdir or adding better examples, but the existing description is not wrong.

[mauke]

PS: Your suggestion of while (defined readdir $dh) is actually wrong. It does not match either of the rewrite rules described above, so no assignment to $_ is added (i.e. the loop body has no access to the current filename anymore). You need to use one of while (readdir $dh), while ($_ = readdir $dh), or while (defined($_ = readdir $dh)); these all do the same thing.

[znik3040]

You are completly right.
while (defined my $item=readdir $dh) {
^^^^^^^^^^^^^ syntax, missed parentheses. it is interpreted as:
while ((defined my $item)=readdir $dh) {
big mistake. defined cannot be modified.

correct version:
while (defined(my $item=readdir $dh)) {

In the same doc is:
As of Perl 5.12 you can use a bare
and later is info, readdir returns into condition info about value is defined. It os done behind, this cause 'defined' is only option.

Then I close this hmmm. stupid issue ;) but for a years I used 'defined'. Thank you for support.

[demerphq]

On Thu, 13 Feb 2025 at 13:47, mauke ***@***.***> wrote: (In old perls before v5.25, an empty hash would yield an empty string

instead, i.e. point 1.) Actually in older perls scalar(%hash) would return a string of the form "$keys/$buckets" (Nowadays exposed as Hash::Util::bucket_ratio()). Later on we special cased if (%hash) to return the number of keys, and later on (perl 5.25 as you said) we made it so that scalar(%hash) did the same thing. Exposing the bucket count had little value, and was a performance problem. cheers, Yves

…

-- perl -Mre=debug -e "/just|another|perl|hacker/"

[mauke]

Oops, looks like I misremembered. Even on perls before v5.25, an empty hash in scalar context returns an integer 0. (Non-empty hashes generate a string as you describe.)