Perl 正则表达式中的 1 和 $1 有什么区别? [英] What is the difference between 1 and $1 in a Perl regex?

问题描述

使用 1 与 $1(如果有的话)有什么区别，或者它们在所有情况下都可以互换.

What is the difference of doing 1 as opposed to $1 if any, or are they interchangeable in all situations.

示例:

s/([a-z]+),afklol/$1,bck/;
#against
s/([a-z]+),afklol/1,bck/;

它们都给出了相同的结果，但有什么不同吗?

They both give the same result but is there any difference?

推荐答案

直接来自 perldoc perlre:

警告 1 vs $1

有些人太习惯写作了诸如:

Some people get too used to writing things like:

$pattern =~ s/(W)/\1/g;

这是一个 RHS 的祖父替代品以避免震惊 sed上瘾者，但这是一个肮脏的习惯进入.那是因为在 PerlThink 中，s///"的右侧是一个双引号字符串.1"在通常的双引号字符串表示一个控制-A.习惯的Unix含义的1"被塞进s///".但是，如果你养成了这样的习惯这样做，你就会进入如果你再添加一个/e"，那就麻烦了修饰符.

This is grandfathered for the RHS of a substitute to avoid shocking the sed addicts, but it’s a dirty habit to get into. That’s because in PerlThink, the righthand side of an "s///" is a double- quoted string. "1" in the usual double-quoted string means a control-A. The customary Unix meaning of "1" is kludged in for "s///". However, if you get into the habit of doing that, you get yourself into trouble if you then add an "/e" modifier.

s/(d+)/ 1 + 1 /eg;        # causes warning under -w

或者如果你尝试去做

s/(d+)/1000/;

你不能通过说来消除歧义{1}000"，而你可以用${1}000".的操作插值不应混淆与匹配操作反向引用.

You can’t disambiguate that by saying "{1}000", whereas you can fix it with "${1}000". The operation of interpolation should not be confused with the operation of matching a backreference.

当然他们的意思是两个不同的s///"左边的东西.

Certainly they mean two different things on the left side of the "s///".

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