Re^4: search of a string in another string with 1 wildcard

In effect, the wildcards could be anywhere.

why in your previous code there is x before $m?

Comment on Re^4: search of a string in another string with 1 wildcard

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Re^5: search of a string in another string with 1 wildcard by roboticus (Chancellor) on Jul 19, 2014 at 22:42 UTC
carolw: The `substr($regex, $i, $m)` tells us to get `$m` characters from the string `$regex` at position `$i`. When you treat `substr()` as a lvalue (i.e., put it on the left side of the equals sign), you're telling perl to replace that substring with what's on the other side. Since you're replacing `$m` characters, with a '.', you'll lose `$m-1` characters (assuming `$m` is larger than 1). The x is a "repeat" operator, so "." x 5 creates a string of five periods. So we're using `'.' x $m` to create a string composed of periods as long as the substring you're replacing, so you don't change the length of the string. ...roboticus When your only tool is a hammer, all problems look like your thumb.	[reply] [d/l] [select]

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Re^5: search of a string in another string with 1 wildcard
by roboticus (Chancellor) on Jul 19, 2014 at 22:42 UTC

carolw:

The substr($regex, $i, $m) tells us to get $m characters from the string $regex at position $i. When you treat substr() as a lvalue (i.e., put it on the left side of the equals sign), you're telling perl to replace that substring with what's on the other side.

Since you're replacing $m characters, with a '.', you'll lose $m-1 characters (assuming $m is larger than 1). The x is a "repeat" operator, so "." x 5 creates a string of five periods. So we're using '.' x $m to create a string composed of periods as long as the substring you're replacing, so you don't change the length of the string.

...roboticus

When your only tool is a hammer, all problems look like your thumb.

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