Re: How do I assign & substitute in one statement?

You can also use parenthesis to impose 'list context', and then combine that with 'map' to get what you want in a single line. The drawback is you have to shoehorn your string into a list, which is a bit artificial. The benefit is this may be a little bit easier to read and understand what is going on for beginning perl programmers.

$sBegin = "hello world";
($sEnd) = map{s/hello/goodbye/g;$_;}($sBegin);
print $sEnd;
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Comment on Re: How do I assign & substitute in one statement? Download Code

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Re: Answer: How do I assign & substitute in one statement? by Roy Johnson (Monsignor) on Nov 13, 2004 at 03:38 UTC
This is not unlike the earlier `for` solution. Using `for` instead of `map` avoids the clunky returning of `$_`: `$sBegin = 'hello world'; s/hello/goodbye/g for ($sEnd = $sBegin); print $sEnd;` [download] Caution: Contents may have been coded under pressure.	[reply] [d/l] [select]
Re: Answer: How do I assign & substitute in one statement? by Roy Johnson (Monsignor) on Nov 13, 2004 at 15:16 UTC
Important thing I overlooked: your solution modifies `$sBegin`! You would need to create a copy before modifying. One way: `($sEnd) = map {s/hello/goodbye/g; $_;} map {$_} ($sBegin);` [download] Another: `($sEnd) = map {s/hello/goodbye/g; $_;} @{[$sBegin]};` [download] A functionally similar solution that doesn't use map: `$sEnd = do {local $_ = $sBegin; s/hello/goodbye/g; $_;};` [download] Of course, creating a copy and working on it is the canonical solution: `(my $sEnd = $sBegin) =~ s/hello/goodbye/g;` [download] ~~This couldn't be used in a combination declaration-assignment, though, while the others could.~~ Caution: Contents may have been coded under pressure.	[reply] [d/l] [select]
Re: Answer: How do I assign & substitute in one statement? by ysth (Canon) on Nov 14, 2004 at 06:45 UTC
Those parentheses do not impose list context. Parentheses never by themselves make any change to the context of what is in them.	[reply]