爲什麼我的程序即使在提示處什麼也不輸入的情況下工作？

Question

我有以下腳本提示輸入用戶名和密碼並檢查它們是否匹配。但是，即使我只是在提示時輸入輸入，但不輸入用戶名或密碼，它也會打印Success!。爲什麼？

#!/usr/bin/perl 

%data = qw(
    javad root 
    ali  1234 
    george password 
); 

print "Please enter your name :\n"; 
$name = <STDIN>; 
chomp($name); 
$pass = $data{$name}; 

#receiving password from user 
#compare it with correct password 
#print the result 
print "What is your password ?\n"; 
$guess = <STDIN>; 
chomp($guess); 

$whileTest = "True"; 
while ($whileTest eq "True") { 
    if ($pass eq $guess) { 
     print "Success!\n"; 
     $whileTest = "False"; 
    } 
    elsif ($pass ne $guess) { 
     print "Wrong password , Try again :\n"; 
     $guess = <STDIN>; 
     chomp($guess); 
    } 
} 

Answer 1

快速修復

$pass = $data{$name} or die "No password for $name"; 

當你輸入空白名稱，密碼是undef後來與空字符串相等於真實的，允許訪問。 use warnings;至少會抱怨這種比較。