Question

我是一个Perl新手，试图找出为什么我的直到条件在基本猜数游戏中从未评估为真。

#!/usr/bin/perl
# Ex 10-1

use warnings;

$num = int(1 + rand 100);
print "\$num is $num\n";                # Debug

print "I'm thinking of a number between 1 and 100. What number is it?\n";
until (chomp($guess = <STDIN>) == $num) {
    print "\$guess is $guess\n";        # Debug
    if ($guess =~ /\A\s*(quit|exit)?\s*\z/) {
        print "Exiting.\n";
        exit 0;
    } elsif ($guess =~ /\D/) {
        print "Invalid input. ";
    } else {
        ($guess < $num) ? (print ("Too low. ")) : (print ("Too high. "));
    }
    print ("Guess again.\n");
}
print ("Congrats! You guessed the number.\n");
exit 0;

以下是执行的一个例子：

$ ./ex1
$num is 4
I'm thinking of a number between 1 and 100. What number is it?
5
$guess is 5
Too high. Guess again.
3
$guess is 3
Too low. Guess again.
4
$guess is 4
Too high. Guess again.

将$ guess与$ num进行比较，当输入4时，$ guess等于$ num，满足测试条件。然而它仍然进入循环并使其达到打印的最终（默认）结果并且太高了。＆＃34;

我还尝试在chomp（$ guess =）周围添加另一组括号，但这并没有帮助。

为什么会这样？

Answer 1

chomp返回删除的字符数，而不是chomp之后的修改后的参数，因此您不应该在循环条件中使用返回值。

chomp(my $guess = <STDIN>);

until ($guess == $num) {
    ...
    chomp($guess = <STDIN>);
}

Perl：直到条件从未评估为真

1 个答案: