perl中的chomp无法按预期工作

问题描述：

我在Perl中发现了chomp的一种奇怪行为，我无法理解为什么chomp会像这样工作.

I found a strange behavior of chomp in Perl and I am unable to comprehend why is chomp is working like this.

以下行无法按预期运行

if ( chomp($str1) eq chomp($str2) )

但是，以下方法可以正常工作

But, the following works fine

chomp $str1;
chomp $str2;
if ( $str1 eq $str2 )

请问您对这种行为有何见解?

Can you please give some insight in this behavior of chomp?

答

chomp修改其参数.它不返回修改后的参数.实际上，第二个示例是您应该如何使用它.

chomp modifies its argument. It does not return a modified argument. The second example is, in fact, how you're supposed to use it.

编辑 :perldoc -f chomp说:

   chomp   This safer version of "chop" removes any trailing string that
           corresponds to the current value of $/ (also known as
           $INPUT_RECORD_SEPARATOR in the "English" module).  It returns
           the total number of characters removed from all its arguments.

perl中的chomp无法按预期工作

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