Perl 为什么使用以下脚本时输出不同？

我的输入文件：

103M:A|PDBID|CHAIN|SEQUENCE
MVLSEGEWQLVLHVWAKVEADVAGHGQDILIRLFKSHPETLEKFDRFKHLKTEAEMKASEDLKKAGVTVLTALGAILKKKGHHEAELKPLAQSHATKHKIPIKYLEFISEAIIHVLHSRHPGNFGADAQGAMNKALELFRKDIAAKYKELGYQG

脚本1：

open(FH,"103m.txt")||die "error";
while(<FH>)
{
    print <FH>;
}

脚本2输出：

103M:A|PDBID|CHAIN|SEQUENCE
MVLSEGEWQLVLHVWAKVEADVAGHGQDILIRLFKSHPETLEKFDRFKHLKTEAEMKASEDLKKAGVTVLTALGAILKKKGHHEAELKPLAQSHATKHKIPIKYLEFISEAIIHVLHSRHPGNFGADAQGAMNKALELFRKDIAAKYKELGYQG

---

这实际上是一个有趣的问题，因为有两个不同的perl概念在起作用

第一个是-

导致从文件中读取一行

因此，如果你：

while ( <FH> ) {

以及：

这将输出：

Loop count 1
$_ is "line 1 
"
Printing <DATA>
line 2
line 3
line 4
line 5

注意-上面没有

chomp

，因此

$\uu

包含换行符

当我们在做这件事的时候，那种形式的

open

并不是一种好的做法。相反，我建议：

open ( my $input, '<', "103m.txt" ) or die $!; 

open（我的$input，'
while ( defined $_ = <FH> ) { 

}

my $line = <FH>;

my @lines = <FH>; 

#!/usr/bin/env perl
use strict;
use warnings;

my $count; 

while ( <DATA> ) {
   print "Loop count ", ++$count,"\n";
   print '$_ is "', $_,"\"\n";
   print "Printing <DATA>\n";
   print <DATA>;
}

__DATA__
line 1 
line 2
line 3
line 4
line 5

Loop count 1
$_ is "line 1 
"
Printing <DATA>
line 2
line 3
line 4
line 5

#!/usr/bin/env perl
use strict;
use warnings;

my $count; 

while ( <DATA> ) {
   print "Loop count ", ++$count,"\n";
   print '$_ is "', $_,"\"\n";
   print 'printing $_',"\n";
   print $_;
}

__DATA__
line 1 
line 2
line 3
line 4
line 5

Loop count 1
$_ is "line 1 
"
printing $_
line 1 
Loop count 2
$_ is "line 2
"
printing $_
line 2
Loop count 3
$_ is "line 3
"
printing $_
line 3
Loop count 4
$_ is "line 4
"
printing $_
line 4
Loop count 5
$_ is "line 5"
printing $_
line 5

open ( my $input, '<', "103m.txt" ) or die $!;