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  perl的join和split问题 - yongjian [ 2005-01-07 06:15 | 487 byte(s)]
Re: perl的join和split问题 - youzhiqingnian [ 2005-01-07 15:22 | 113 byte(s)]
Re: perl的join和split问题 - yongjian [ 2005-01-07 23:10 | 98 byte(s)]
Re: perl的join和split问题 - youzhiqingnian [ 2005-01-09 14:35 | 253 byte(s)]
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$_="abc abc abcabc";
@a=split;
print "the @a is @a
";
$b=join /s+/,@a;
print "the $b is $b
";
output:
---------------------------
the @a is abc abc abcabc
the $b is abc1abc1abcabc
---------------------------
上面是一个小程序，只是想熟悉一下join和split,但是当用s+来join原来的string的时候，space全变成了１了。很奇怪，哪里错了呢？
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Assume we could join a group of things by an pattern, how did Perl do that? I think that's why function join does not accept pattern as its first argument - it is ambigous.
see perldoc -f join and perldoc -f split
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----
It's better to burn out than to fade away...
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用一次join很难实现，如果不是完全不可能的话。反正我还想
不出来用一次join如何在不同位置上用不同的值连接。join对
后面跟的第一个参数只evaluate一次。
所以你要想得到原来的值，最简单的办法就是保留split以前
的变量。
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