Re: Replacing substrings in a string

[David Given <dg@...>]

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Philippe Lhoste wrote:
[...]
> IIRC, the original message was concerned only by the percent sign (or so
> I understood), so I addressed only this case.

Actually, I just mentioned % as an example; I wanted a generic solution.

[...]
>> I suggest that (not tested) :
>>
>> pattern = string.gsub(generatedPattern, "(%c)", "%1")
>> newString = string.gsub(oldString, pattern, replace)
[...]
> I think that the control characters are those below Ascii 32, not the
> chars listed above.

It's grotesque, but:

pattern = string.gsub(generatedPattern, "(.)", "%%%1")
newString = ...

...might just work. It is, however, an awful lot of work --- I hate to think
how many memory allocations are required to do that.

Will it always be the case that the above will be faster than any in-Lua
looping code, due to Lua's dispatch overheads?

- --
+- David Given --McQ-+ "If you're up against someone more intelligent
|  dg@cowlark.com    | than you are, do something insane and let him think
| (dg@tao-group.com) | himself to death." --- Pyanfar Chanur
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