[英]problem passing 0 as command-line argument
I've just noticed a strange behavior of perl5 (5.10.0) when I passed 0 as the command-line argument to a script, which assigns it to a variable. 我刚刚注意到perl5(5.10.0)的奇怪行为,当我将0作为命令行参数传递给脚本时,它将其赋值给变量。 The problem I encountered is that if I give the variable a default value in my script which is greater than 0, say 1, then I can pass any value except 0 to override the default value 1.
我遇到的问题是,如果我在我的脚本中给变量一个大于0的默认值,比如1,那么我可以传递除0之外的任何值来覆盖默认值1。
Following is the code of 'test.pl' I used: 以下是我使用的'test.pl'的代码:
#!/usr/bin/perl -w
my $x = shift || 1;
print "x is $x\n";
Following are the commands and ouputs when I tested it: 以下是我测试时的命令和输出:
$ ./test.pl 2
x is 2
$ ./test.pl 0
x is 1
I'd appreciate if anyone can shed light on this. 如果有人能说清楚,我会很感激。 Thanks.
谢谢。 wwl
WWL
If you want $x
to have the value of "1" in case no argument is provided, use this: 如果您希望
$x
在没有提供参数的情况下具有值“1”,请使用:
my $x = shift // 1;
From perldoc perlop: 来自perldoc perlop:
"//" is exactly the same as "||", except that it tests the left hand side's definedness instead of its truth.
“//”与“||”完全相同,只是它测试左侧的定义而不是它的真实性。
Note: the defined-or operator is available starting from 5.10 (released on December 18th, 2007) 注意:定义的或运算符从5.10开始(2007年12月18日发布)
The expression shift || 1
表达式
shift || 1
shift || 1
chooses the rhs iff the lhs evaluates to "false". shift || 1
选择rhs iff lhs评估为“false”。 Since you are passing 0, which evaluates to "false", $x
ends up with the value 1. 由于您传递0(计算结果为“false”),因此
$x
最终得到值1。
The best approach is the one in eugene's and Alexandr's answers, but if you're stuck with an older perl, use 最好的方法是在eugene和Alexandr的答案中,但如果你坚持使用旧的perl,请使用
my $x = shift;
$x = 1 unless defined $x;
Because you have a single optional argument, you could examine @ARGV
in scalar context to check how many command-line arguments are present. 因为您有一个可选参数,所以可以在标量上下文中检查
@ARGV
以检查存在多少个命令行参数。 If it's there, use it, or otherwise fall back on the default: 如果它在那里,使用它,或以其他方式回到默认值:
my $x = @ARGV ? shift : 1;
use defined-or operator, because string "0" is defined, but not evaluated as true. 使用defined-或operator,因为字符串“0”已定义,但未评估为true。
#!/usr/bin/perl -w
my $x = shift // 1;
print "x is $x\n";
In this line my $x = shift || 1;
在这一行
my $x = shift || 1;
my $x = shift || 1;
the shift failed the test and therefore the conditional logical OR ||
移位未通过测试,因此条件逻辑OR
||
was executed and assigned 1 to it... as per the page on shift the value was 0 which implies empty array.... 执行并分配给它...根据移位页面的值,该值为0,这意味着空数组....
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.