为什么 `$str1 =~ "foo"` 被识别为 `$str1 =~ m/foo/`(而不是语法错误)? [英] Why is `$str1 =~ "foo"` recognized as `$str1 =~ m/foo/` (and not a syntax error)?

问题描述

我最近看到了一个不熟悉 Perl 的人写的代码.他想比较两个字符串是否相等，但不知道eq 操作符，所以他使用 =~ 像这样:

my $str1 = 'foobar';我的 $str2 = 'bar';如果( $str1 =~ $str2 ){print "字符串相等\n";}

另一个片段是

if ( $str1 =~ "foo" ) {打印 "字符串等于 'foo'\n";}

当然它应该简单地读取 $str1 eq $str2 和 $str1 eq "foo" 以避免误报.

我通过 Deparse 运行代码，它说一切正常:

$ perl -MO=Deparse -e '使用严格；使用警告；我的 $str1="foobar";我的 $str2="bar";$str1 =~ $str2;$str1 =~ "bar";'使用警告；使用严格；我的 $str1 = 'foobar';我的 $str2 = 'bar';$str1 =~/$str2/;$str1 =~/bar/;-e 语法正常

我浏览了文档，但根据我的理解情况如下:

• 一般语法是m/pattern/.
• 要么使用 m 和您选择的分隔符而不是 /(但请注意 ' 和? 有特殊含义)
• 或者去掉m，但是分隔符必须是/.

但显然 Perl 将 $str1 =~ "foo" 理解为 $str1 =~ m/foo/ 尽管没有 m 存在.这是为什么?我原以为这是一个语法错误.

解决方案

我原以为这是一个语法错误.

在perlop中引用=~的文档，><块引用>

如果正确的参数是表达式而不是搜索模式、替换或音译，则在运行时将其解释为搜索模式.

<小时><块引用>

但显然 Perl 将 $str1 =~ "foo" 理解为 $str1 =~ m/foo/ 尽管没有 m存在.这是为什么?

为什么不呢?如果 RHS 上没有匹配、替换或音译运算符，我想不出不让 =~ 暗示匹配运算符的理由.我会用

$s =~/foo/

结束

$s =~ "foo"

但我用过

$s =~ $re

特别是当$re的值是由qr//编译的模式时.

I recently saw code from someone who's not familiar with Perl. He wanted to compare two strings for equality but didn't know about the eq operator, so he used =~ like this:

my $str1 = 'foobar';
my $str2 = 'bar';
if ( $str1 =~ $str2 ) {
    print "strings are equal\n";
}

Another snippet was

if ( $str1 =~ "foo" ) {
    print "string equals 'foo'\n";
}

Of course it should simply read $str1 eq $str2 and $str1 eq "foo" to avoid false-positives.

I run the code through Deparse and it said everything is ok:

$ perl -MO=Deparse -e 'use strict; 
                       use warnings; 
                       my $str1="foobar"; 
                       my $str2="bar"; 
                       $str1 =~ $str2; 
                       $str1 =~ "bar";'
use warnings;
use strict;
my $str1 = 'foobar';
my $str2 = 'bar';
$str1 =~ /$str2/;
$str1 =~ /bar/;
-e syntax OK

I looked through the docs, but from my understanding the situation is as follows:

• The general syntax is m/pattern/.
• Either use m and a delimiter of your choice instead of / (but be aware that ' and ? have special meaning)
• Or leave off the m but then the delimiter must be /.

But apparently Perl understands $str1 =~ "foo" as $str1 =~ m/foo/ although no m is present. Why is that? I'd expected that to be a syntax error.

解决方案

I'd expected that to be a syntax error.

Quoting the documentation for =~ in perlop,

If the right argument is an expression rather than a search pattern, substitution, or transliteration, it is interpreted as a search pattern at run time.

---

But apparently Perl understands $str1 =~ "foo" as $str1 =~ m/foo/ although no m is present. Why is that?

Why not? I can't think of a reason not to have =~ imply a match operator if there's no match, substitution or transliteration operator on its RHS. I'd use

$s =~ /foo/

over

$s =~ "foo"

but I have used

$s =~ $re

Especially when the value of $re is a pattern compiled by qr//.

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