It's hard to see s/this/that/ as an opperator! Because, err ... it looks like an expression

s/this/that/ *is* an expression. The expression consists of the substitution operator and its two operands (the search pattern and the replacement expression).

$text =~ s/this/that/ is also an expression. The expression consists of the binding operator (=~) and its two operands, the value to bind ($text) and the previously mentioned expression.

The binding operator does not return its LHS. It returns its RHS. As for the substitution operator, it does not return the variable to which it is bound.

the following which prints 7 as expected.

I'm not sure what you are saying. Are you saying $var = 3+4 returns the result of 3+4, so $text =~ s/this/that/ should return the result of s/this/that? That's wrong for two reasons.

I didn't say it's not possible for an operator to return it's LHS. As you've shown, scalar assignment returns its left-hand side. There's also boolean operators. They return either their LHS or RHS operands.

Better solution: 

foreach (keys %{$ref_cookie}) {
    push @{$self->{cookies}}, /^-(.*)/, $ref_cookie->{$_};
}

[download]

The match operator returns what it captures.

Update: Remove trailing paren

In reply to Re^3: Inline substitution regex by ikegami
in thread Inline substitution regex by bradcan

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