我继续在use strict上使用strict来获取初始化错误,并且无法找出原因。我为所有变量设置了范围 - 并且代码运行。只是不明白丑陋的错误。
2/15/2002 Joe 155 2/15/2002 Mike 108 2/15/2002 Pete 209 2/22/2002 Joe 158 2/22/2002 Mike 99 2/22/2002 Pete 163 3/1/2002 Joe 172 3/1/2002 Mike 125
#!/usr/bin/perl -w
our %dates;
foreach my $line (<>) {
chomp $line;
my ($this_date, $this_name, $this_score) = split /\s+/, $line;
my ($record_name, $record_score) = split /\|/, $dates{$this_date};
if ($this_name && $this_score) {
if ($this_score > $record_score) {
$dates{$this_date} = join "|", ($this_name, $this_score);
}
}
}
foreach my $date (keys %dates) {
my ($name, $score ) = split /\|/, $dates{$date};
print " The high_scored for $date was $name with $score\n";
shortcasper@shortcasper-laptop:~/perl$ ./hash_bowl bowl_linux Use of uninitialized value in split at ./hash_bowl line 8, line 7. Use of uninitialized value $record_score in numeric gt (>) at ./hash_bowl line 10, line 7. Use of uninitialized value in split at ./hash_bowl line 8, line 7. Use of uninitialized value $record_score in numeric gt (>) at ./hash_bowl line 10, line 7. The high_scored for 3/1/2002 was Joe with 172 The high_scored for 2/15/2002 was Pete with 209 shortcasper@shortcasper-laptop:~/perl$
答案 0 :(得分:6)
您应该use warnings
而不是-w
。
它抱怨的原因是,当您第一次遇到某一天时,$dates{$this_date}
为undef
(因为它从未设置过)。拆分会给您一个警告并生成$record_name
和$record_score
undef
(在您将$this_score
与$record_score
进行比较时会发出第二次警告)。代码可以正常工作,因为数字undef
被认为是0,但它会生成警告。
一个简单的解决方法是使用$dates{$this_date} || '|0'
代替。这为新日期提供了默认值,将$record_name
设置为空字符串,将$record_score
设置为0:
use strict;
use warnings;
our %dates;
foreach my $line (<DATA>) {
chomp $line;
my ($this_date, $this_name, $this_score) = split /\s+/, $line;
my ($record_name, $record_score) = split /\|/, $dates{$this_date} || '|0';
if ($this_name && $this_score) {
if ($this_score > $record_score) {
$dates{$this_date} = join "|", ($this_name, $this_score);
}
}
}
foreach my $date (keys %dates) {
my ($name, $score ) = split /\|/, $dates{$date};
print " The high_scored for $date was $name with $score\n";
}
__DATA__
2/15/2002 Joe 155
2/15/2002 Mike 108
2/15/2002 Pete 209
2/22/2002 Joe 158
2/22/2002 Mike 99
2/22/2002 Pete 163
3/1/2002 Joe 172
3/1/2002 Mike 125
但您应该阅读Perl Data Structures Cookbook,并考虑使用复杂的数据结构,而不是只需将join
和split
数据存储在哈希值中。
答案 1 :(得分:2)
在我看来,您的哈希%dates
已初始化但其值不是。您的代码在实际设置之前尝试使用$dates{$this_date}
的值(稍后会发生几行)。
答案 2 :(得分:0)
%date是一个空哈希。所以$ dates {$ this_date}正在变成“undef”
改变你的内心状况,使其发挥作用;当没有旧记录时,它将插入日期哈希。
if ( !(defined $record_score) || $this_score > $record_score) {
$dates{$this_date} = join "|", ($this_name, $this_score);
}