perl - 分配给变量的未定义值。

Question

5 > 6 的返回值是 undef。但是，当我将 5 > 6 的值分配给定义该变量的变量时。如何使用失败时解析为 undef 的比较运算符传递语句的值？

#!/usr/bin/perl
use strict ;
use warnings;
print'Five is more than six ? ',      5 >  6, "\n";
print 'Five is less than six ? ' , 5 < 6 , "\n";
my $wiz = 5 > 6 ;

if (defined($wiz)) {
    print '$wiz is defined' ;
    } else {
    print '$wiz is undefined' ;
}


$ ./lessthan
Five is more than six ?
Five is less than six ? 1
$wiz is defined

Answer 1

How do I pass the value of a statement using a comparison operator that resolves to undef on failure?

一般来说，Perl 不 promise 从其运算符返回任何特定的真值或假值，并且 <也不异常(exception)。如果你想要特定的值，你将需要像

$boolean ? $value_for_true : $value_for_false

所以

my $wiz = 5 > 6 ? 1 : undef;

如果你只关心你得到的 false 的值，你也可以使用下面的:

my $wiz = 5 > 6 || undef;

事实证明这两个选项是等价的，但这并不能保证。

---

The return value of 5 > 6 is undef. However when I assign the value of 5 > 6 to a variable

那不是真的。该变量被分配了一个定义的值，因为 5 > 6评估为定义的值。虽然 Perl 没有指定它从它的运算符返回什么假值，但它通常返回标量 sv_no。 ，这是一个 dualvar，在被视为字符串时显示为空字符串，在被视为数字时显示为零。

$ perl -wE'say "> ", "".( undef )'
Use of uninitialized value in concatenation (.) or string at -e line 1.
>

$ perl -wE'say "> ", "".( "" )'
>

$ perl -wE'say "> ", "".( 0 )'
> 0

$ perl -wE'say "> ", "".( 5>6 )'
>                                     # Behaves as ""

$ perl -wE'say "> ", 0+( undef )'
Use of uninitialized value in addition (+) at -e line 1.
> 0

$ perl -wE'say "> ", 0+( "" )'
Argument "" isn't numeric in addition (+) at -e line 1.
> 0

$ perl -wE'say "> ", 0+( 0 )'
> 0

$ perl -wE'say "> ", 0+( 5>6 )'
> 0                                   # Behaves as 0