Re: regexp operator -- same mistake over and over again

Makes sense to me. Every other operator with the 'equals' in it has the 'equals' last, e.g. += or *= (Otherwise $foo =+ 20 would be parsed as $foo = +20)

The =~ is the opposite and that makes it confusing.

Comment on Re: regexp operator -- same mistake over and over again

Replies are listed 'Best First'.
Re: Re: regexp operator -- same mistake over and over again by Anonymous Monk on Jan 23, 2004 at 12:23 UTC
by analogy with += `$foo ~= s/a/b/;` [download] should be equivalent to `$foo = $foo ~ s/a/b/;` [download] so maybe ~ should be a binary operator that returns the result of the substitution whereas ~= would apply the substitution to $foo	[reply] [d/l] [select]