What does this expression do in Perl regex?

Question

$_ = ~m [($substr)]g;

I have three questions:

1. What's the difference between ~m[] and ~m//?
2. Why if deleted () the results would be different?
3. What does it mean by the g at the end?

Answer 1

1. There is no difference between /.../, m/.../ and m[...]. You can use the delimiter of your choice when you specify the m. /.../ is standard, but sometimes it's more readable to use something else. For example, compare

   /^http:\/\//
   

   with

   m[^http://]
   

2. () are captures. They capture the text matched by the pattern within. You can access the captured text of the first capture via $1, the second via $2, etc. In list context, the match operator returns the captured strings.

   $ perl -E'say for 'abcd' =~ /(.).(.)/;'
   a
   c
   

3. m//g is used to find all the matches. m// is documented in perlop.

   $ perl -E'say for "abc" =~ /(.)/;'
   a
   
   $ perl -E'say for "abc" =~ /(.)/g;'
   a
   b
   c
   

Note that $var = ~m[...] and $var =~ m[...] are very different, and that you surely meant to use the latter. =~ followed by a match operator specifies the variable against which the match operator is matching.

Answer 2

1. The ~m[] and ~m// are same, and they are boundary symbols.

2. Scalar variable $substr is embed into parentheses, it means that you can get matched value into parentheses through print $1.

3. The symbol g means global match.