シンプルなPerlのforループエラー

Question

for ($i=1; $i<=3; $i++) { 
    if (@user$i[3] >= 4.0) { 
     print "first"; 
    } elsif (@user$i[3] < 4.0 and @user$i[3] >= 2.5) { 
     print "second"; 
    } else { 
     print "third"; 
    } 
} 

それは私がコード

以下同じ機能は、あなたが私を反復処理する

if (@user1[3] >= 4.0) { 
    print "first"; 
} elsif (@user1[3] < 4.0 and @user1[3] >= 2.5) { 
    print "second"; 
} else { 
    print "third"; 
} 

if (@user2[3] >= 4.0) { 
    print "first"; 
} elsif (@user2[3] < 4.0 and @user2[3] >= 2.5) { 
    print "second"; 
} else { 
    print "third"; 
} 

if (@user3[3] >= 4.0) { 
    print "first"; 
} elsif (@user3[3] < 4.0 and @user3[3] >= 2.5) { 
    print "second"; 
} else { 
    print "third"; 
} 

Answer 1

使用の参照をいくつかの助けを与えることができるしたい

を動作しません。異なる配列。

perldoc perlref - Perlのリファレンス

perldoc perldsc - perlのデータ構造クックブック

#/usr/bin/env perl 

use strict; 
use warnings; 

my @user1 = (1, 2, 3, 4); 
my @user2 = (4, 3, 1, 2); 
my @user3 = (2, 1, 4, 3); 

my $data = [ \@user1, \@user2, \@user3 ]; 
my $field = 3; # Fourth field, perl is 0-based           

for my $users (@$data) { 
    # $users->[...] instead of $users[...] because $users is an array reference    
    if ($users->[$field] >= 4.0) { 
     print "first"; 
    } elsif ($users->[$field] < 4.0 and $users->[$field] >= 2.5) { 
     print "second"; 
    } else { 
     print "third"; 
    } 
    print "\n"; 
} 

出力

first 
third 
second