Question

我真的很困惑如何将变量传递给不同的语句。我的代码是：

use warnings;
use strict;
use feature qw(say);

say "Please enter the first sequence:";
my $sequence1 = <STDIN>;
$sequence1 = chomp $sequence1;
say "Please enter the second sequence:";
my $sequence2 = <STDIN>;
$sequence2 = chomp $sequence2;

if (length $sequence1 < length $sequence2){
        my $sequence2_new = substr $sequence2, length $sequence1;
}

my @sequence1 = split(',', $sequence1);
my @sequence2 = split(',', $sequence2_new);
my $element = scalar @sequence1;

my $num = 0;
for ($a = 0; $a < $element; $a++){
        if ($sequence1[$a] = $sequence2[$a]){
                $num++;
        }
}

my $score = $num % length $sequence2;
say "The alignment score is: $score";

在这种情况下，此程序将返回Global symbol "$sequence2_new" requires explicit package name at alignment_sequence.pl line 19.错误。如果我移动＆＃34;我的＆＃34;第14行if语句前面的声明，如my $sequence2_new;，它会给我Use of uninitialized value $sequence2_new in split at alignment_sequence.pl line 20, <STDIN> line 2.警告。

Answer 1

您需要在my $sequence2_new语句之外声明if 。如你所知，变量的生命在if块

结束时结束

另请注意

$sequence1 = chomp $sequence1

错了。它会将$sequence1设置为chomp删除的字符数 - 可能为1或0.您只想

chomp $sequence1

您还有if ( $sequence1[$a] = $sequence2[$a] ) { ... } 分配。大概你想要比较器eq吗？

以下是我认为您的代码应该如何显示的内容，但如果$sequence2长于$sequence1，我完全不能确定它是否已切断use strict; use warnings; use feature qw(say); print "Please enter the first sequence: "; chomp ( my $sequence1 = <> ); print "Please enter the second sequence: "; chomp ( my $sequence2 = <> ); my $sequence2_new; if ( length $sequence1 < length $sequence2 ) { $sequence2_new = substr $sequence2, length $sequence1; } my @sequence1 = split /,/, $sequence1; my @sequence2 = split /,/, $sequence2_new; my $num = 0; for my $a ( 0 .. $#sequence1 ) { ++$num if $sequence1[$a] eq $sequence2[$a]; }的开头。这看起来并不合适，但我无法确定

Dim LastRow As Long
LastRow = Sheets("sheet1").Range("A" & Rows.Count).End(xlUp).Row

如何将变量传递给不同的if语句

1 个答案: