glob in scalar context

atiking

起因是我看了这篇
http://coding.derkeiler.com/Archive/Perl/comp.lang.perl.misc/2007-06/msg01421.html

于是我做了些修改

1. ~/temp $ touch foo.1 foo.2 foo.3
   
2. ~/temp $ perl -le'
   
3. for (1..10) {
   
4. if (glob("foo.$_")) {
   
5. print "yes"
   
6. } else {
   
7. print "no"
   
8. }
   
9. }
   
10. '
    
11. yes
    
12. no
    
13. yes
    
14. no
    
15. yes
    
16. no
    
17. yes
    
18. no
    
19. yes
    
20. no
    

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为什么呢，以我的理解结果应该是

1. 
   
2. yes
   
3. yes
   
4. yes
   
5. no
   
6. no
   
7. no
   
8. no
   
9. no
   
10. no
    
11. no
    

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jason680

起因是我看了这篇


于是我做了些修改为什么呢，以我的理解结果应该是
atiking 发表于 2010-12-16 17:32

if (glob("foo.$_")) {
改   
if ( -f "foo.$_") { 或 if ( -e "foo.$_") {

atiking

if (glob("foo.$_")) {
改   
if ( -f "foo.$_") { 或 if ( -e "foo.$_") {
jason680 发表于 2010-12-16 18:19

这么改没错

只是想知道glob为什么会有那样的结果

zhlong8

按文档中说 scalar context 中 glob 返回个 iterator，最后会返回个 undef 表示结束，教你怎么查检

1. for (1..10) {
   
2. if ($name = glob("foo.$_")) {
   
3. print "yes ", $name;
   
4. } else {
   
5. print "no"
   
6. }
   
7. }

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兰花仙子

起因是我看了这篇


于是我做了些修改为什么呢，以我的理解结果应该是
atiking 发表于 2010-12-16 17:32

    From "perldoc -f glob":

    In scalar context, glob iterates through ... filename expansions,
    returning undef when the list is exhausted.

So glob() is returning the file pattern itself on the first call, as
Charles described, and then undef on the second call to indicate that
the list is complete. The cycle continues on the third call and so on.

To avoid this effect, call glob() in list context, like this perhaps:

  for (1..10) {
    if (my @x = glob "foo.$_") {
      print "@x\n";
      print "yes";
    }
    else {
      print "@x\n";
      print "no";
    }
    print "-------------";
  }

HTH,

Rob

atiking

按文档中说 scalar context 中 glob 返回个 iterator，最后会返回个 undef 表示结束，教你怎么查检
zhlong8 发表于 2010-12-16 20:41

那这个iterator是根据什么返回的呢
现在来看似乎和glob中的pattern没什么关系？

兰花仙子

那这个iterator是根据什么返回的呢
现在来看似乎和glob中的pattern没什么关系？
atiking 发表于 2010-12-17 09:28

    There weren't any  matches with files and no wildcards
in the glob pattern so the pattern itself is returned *.  If
there had been a wildcard in the pattern, then nothing
would have been returned.

perl -E "say glob('foo.3')"   <--- returns foo.3 *
perl -E "say glob('foo?3)"   <--- doesn't return anything

   *  You'd have to use bsd_glob in order to  not return
       the pattern if no files match, eg:
              use File::Glob ':glob';
              say bsd_glob('foo.3',GLOB_ERR);

      Otherwise, if there had been a file "foo.3" with an
      actual '.' in the filename, glob('foo.3') would also
      return that filename.

Perl uses File::Glob for globbing since 5.6.1 and  only
these metacharacters are recognized:

      \       Quote the next metacharacter
      []      Character class
      {}      Multiple pattern
      *       Match any string of characters
      ?       Match any single character
      ~       User name home directory

For more detail: perldoc File::Glob

Perhaps, you were thinking '.' was also a glob wildcard
as it  would be in a regex pattern...

--
Charles DeRykus

atiking

回复 5# 兰花仙子
也就是说在scalar的环境下，如果在list没有结束的情况下
再次glob中的pattern无效
glob会继续返回上次pattern的list中的下一个结果

atiking

回复  兰花仙子
也就是说在scalar的环境下，如果在list没有结束的情况下
再次glob中的pattern无效
glob ...
atiking 发表于 2010-12-17 09:38

我又升级了一下测试脚本，验证了上述的推论

1. 
   
2. ~/temp $ touch foo.1 foo.11 foo.2 foo.3 foo.4 foo.5 foo.6
   
3. ~/temp $ cat test.pl
   
4. #/usr/bin/perl
   
5. use strict;
   
6. 
   
7. for (1..10) {
   
8.     print "$_\t";
   
9.    
   
10.     if (my file glob("foo.*$_")) {
    
11.         print "$file\tyes"
    
12.     } else {
    
13.         print "no"
    
14.     }
    
15.     print "\n";
    
16. }
    
17. ~/temp $ perl test.pl
    
18. 
    
19. 1    foo.1    yes
    
20. 2    foo.11  yes
    
21. 3    no
    
22. 4    foo.4    yes
    
23. 5    no
    
24. 6    foo.6    yes
    
25. 7    no
    
26. 8    no
    
27. 9    no
    
28. 10  no
    

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