具有全局替换的perl regexp中的奇怪行为

Question

可以解释为什么这个小的perl脚本的输出是"foofoo"(而不是"foo")？

#!/usr/bin/perl -w 
my $var="a";
$var=~s/.*/foo/g;
print $var."\n";

没有g选项它可以工作,虽然它会,但为什么全局选项匹配模式两次？

在bash中,输出是"foo",如预期的那样

echo "a"|sed -e "s/.*/foo/g" 

任何解释将不胜感激.

Answer 1

首先.*匹配a,然后匹配a之后的空字符串.也许你想要.+？

Answer 2

如果你尝试,它会更有趣

$var=~s/.*?/foo/g;

你会得到

foofoofoo

的？修饰符匹配1或0次.如果你删除g,你会得到

fooa

因为它只会替换空字符串,而是它找到的第一个字符串.我爱perl.

Answer 3

这是因为你使用的是.*代替.+

*修饰符告诉正则表达式引擎匹配(并在您的示例中替换)字符串"a",然后是零长度字符串(并替换它).

您可以在示例代码中使用此正则表达式对此进行测试:

$var=~s/(.*)/<$1>/g;

然后,您将看到此输出:

"<a><>"

Answer 4

如果您在代码中添加re:

use re 'debug';

你会看到正则表达式成功匹配两次:

Compiling REx `.*'
size 3 Got 28 bytes for offset annotations.
first at 2
   1: STAR(3)
   2:   REG_ANY(0)
   3: END(0)
anchored(MBOL) implicit minlen 0
Offsets: [3]
        2[1] 1[1] 3[0]
Matching REx ".*" against "a"
  Setting an EVAL scope, savestack=5
   0 <> <a>               |  1:  STAR
                           REG_ANY can match 1 times out of 2147483647...
  Setting an EVAL scope, savestack=5
   1 <a> <>               |  3:    END
Match successful!
Matching REx ".*" against ""
  Setting an EVAL scope, savestack=7
   1 <a> <>               |  1:  STAR
                           REG_ANY can match 0 times out of 2147483647...
  Setting an EVAL scope, savestack=7
   1 <a> <>               |  3:    END
Match successful!
Matching REx ".*" against ""
  Setting an EVAL scope, savestack=7
   1 <a> <>               |  1:  STAR
                           REG_ANY can match 0 times out of 2147483647...
  Setting an EVAL scope, savestack=7
   1 <a> <>               |  3:    END
Match possible, but length=0 is smaller than requested=1, failing!
                            failed...
Match failed
foofoo
Freeing REx: `".*"'