[英]Perl xor returns unexpected result
What's going on here? 这里发生了什么? Why isn't $c == 1 ?
为什么不是$ c == 1?
$ cat y.pl
#!/usr/bin/perl
$a = 0;
$b = 1;
$c = $a xor $b;
print "$a ^ $b = $c\n";
$ ./y.pl
0 ^ 1 = 0
Always use 一直用
use strict;
use warnings qw( all );
It would have detected your precedence issue. 它会检测到您的优先问题。
Useless use of logical xor in void context
Specifically, 特别,
$c = $a xor $b;
means 手段
($c = $a) xor $b;
The operators and
, or
, not
and xor
have very low precedence. 运算符
and
, or
, not
和xor
优先级非常低。 This makes and
, or
and not
very useful for flow control. 这使得
and
, or
和not
流控制是非常有用的。
... or die ...;
... or next;
xor
doesn't short-circuit, so it's not useful for flow control. xor
不会短路,因此对流量控制没有用。 xor
was created for an entirely different reason. xor
是出于完全不同的原因而创建的。 Like or
, xor
is a logical operator. 喜欢
or
, xor
是一个逻辑运算符。 That means considers the truth of its operand as whole instead of doing bit-by-bit comparisons. 这意味着将整个操作数的真实性视为整体而不是逐位比较。 I don't believe this is the operation you want.
我不相信这是你想要的操作。
Fix: 固定:
$c = $a ^ $b;
Low
Precedence
Operation Bitwise Logical Logical
---------- ---------- ---------- ----------
NOT ~ ! not
AND & && and
OR | || or
OR* N/A // N/A
XOR ^ N/A xor
OR* - Like OR, but only undefined is considered false.
Precedence of the assignment is higher than xor. 赋值的优先级高于xor。 Just write:
写吧:
$c = ($a xor $b);
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