开发者

Perl - Use of uninitialized value?

So I'm trying to run this code...

my $filePath = $ARGV['0'];
if ($filePath eq ""){
    print "Missing argument!";
}

It should check the first command line argument, and tell me if its empty, but it returns this error and I can not figure out why:

Use of uninitialized value $filePath in string eq at program.pl line 19.

What am I d开发者_运维百科oing wrong?


Just check to see if $ARGV[0] is defined

#!/usr/bin/perl
use strict;
use warnings;

if(!defined $ARGV[0]){
    print "No FilePath Specified!\n";
}

This will print "No FilePath Specified!\n" if there was none passed command line.

The problem you are running into, is you are setting $filePath to an undefined value. Warnings is complaining because you've then tried to compare an undefined value to "". Warnings thinks that is worth telling you about.

I used my example to show a clean way of checking if something is defined, but technically for this, you could also just do:

if(!@ARGV){
    print "No FilePath Specified!\n";
}


Empty and uninitialized are not the same thing. You can check if a variable is initialized with the defined operator, like for example:

if ((!defined $filePath) || ($filePath eq "")) {
 # $filePath is either not initialized, or initialized but empty
 ...
}

I'm pretty sure you meant this:

my $filePath = $ARGV[0];

(without the quotes)


Alternative answer is to set a default value if it is not defined:

my $filePath = $ARGV[0] // '';


EDIT: As @Andrew pointed out, that's not same, as it will fail with filename "0"

Also, instead of

if ((!defined $filePath) || ($filePath eq "")) { ...

as @Mat wrote. you can use simpler

if(!$filePath) { ...

which does exactly the same

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜