regex - Perl regex extracting a match using braces

Question

I tested the following code

#! /usr/bin/perl 

use strict;
use English;

#this code extracts the current scripts filename 
#by removing the path from the filepath

my $Script_Name = $PROGRAM_NAME;

${Script_Name} =~ s/^.*\\//; #windows path

#${Script_Name} =~ s/^.*\///; #Unix based path

print $Script_Name;

and i don't understand why these braces extract the match without using a /r modifier. can anyone explain why and how this works or point me to some documentation?

Answer 1

你开始有点糊涂了！

牙套没什么区别。${Script_Name}与 相同$Script_Name。

您的代码首先将脚本文件的整个路径从 复制$PROGRAM_NAME到$Script_Name.

然后替换删除所有内容，包括最后一个反斜杠，只留下文件名。

如果您想修改一个字符串并将修改结果放入另一个字符串，则将使用/r修饰符，因此您可以一步编写代码

$Script_Name = $PROGRAM_NAME=~ s/^.*\\//r