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Test condition for pattern match

Please consider following code snippet.  The value of file_path is ..\..\..\src\app\test\filename.c.  What is
[ \  /   \  \ ].  Note:  There is no whitespace inside brackets.  I put whitespace so contents inside parentheses is visible to the experts.  Is the if condition true?
paths_hash is a hash variable.  build_path is M:\view_name\lib\subdir\target\app\test
What does the statement inside if condition do?

if ( $file_path !~ m/[\  /  \  \  ]/ )
{
    $paths_hash{$build_path} = 1;
}
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naseeam
Asked:
naseeam
4 Solutions
 
Adam314Commented:
The if statement is looking to see if $file_path contains either a forward slash "/" or a backslash "\".
So when $file_path = "..\..\..\src\app\test\filename.c", the if condition is true.
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FishMongerCommented:
Adam, I think you meant the opposite.

The if statement is looking to see if $file_path does NOT contain either a forward slash "/" or a backslash "\".

!~ vs =~ operator
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naseeamAuthor Commented:
So forward and backward slash can be in any position.  It doesn't matter that there are two dots before the backward slash.  

Thanks.
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FishMongerCommented:
Yes, the forward or backslash could be at any point in the path.  If either are found, then the result of the conditional is false.

Both of these conditionals will have the same results

if ( $file_path !~ m/[\  /  \  \  ]/ )

unless ( $file_path =~ m/[\  /  \  \  ]/ )
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evilrixSenior Software Engineer (Avast)Commented:
The [] is a character class, it mean match ANY of the chars inside it. In this case the chars are \ and / (both escaped) so it'll match either. The != negates this logic so, as stated above, it means don't match.

So the code means...

if no match is found for \ or / in $file_path then $paths_hash{$build_path} equals 1.

More info on character classes: -

http://www.regular-expressions.info/charclass.html

-Rx.

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mjcoyneCommented:
In summation, $paths_hash{$build_path} is set to equal 1 if the variable $file_path does not contain any forward or backward slashes.
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