Re: [Boost-users] help replace_all_regex and format_literal

From: Pavol Droba (droba_at_[hidden])
Date: 2007-03-15 05:37:47

Hi,

llwaeva_at_[hidden] wrote:
> Hi there,
>   I found that the behavior of replace_all_regex is weird while specifing format flag as format_literal. 
> For example, I would like to replace %f and %g in a source with some
> text respectively. The code is 
> 
> string source = "#f = %f \r\n #g = %g \r\n";
> string text1("<footi_at_[hidden]>");
> string text2("""name""<name_at_email.101.pi>");
> 
> replace_regex(source, boost::regex("%f"), text1, boost::regex_constants::format_literal);
> replace_regex(source, boost::regex("%g"), text2, boost::regex_constants::format_literal);
> 
> The result is fine. However, I would like to use replace_all_regex instead. i.e.
> 
> replace_all_regex( source, boost::regex("(%f|%g)"), string("(?1 loofon_at_[hidden])(?2 ""name""<name_at_email.101.pi>"), boost::regex_constants::format_literal);
> 
> However, because of format_literal, what I get is
> 
> #f = (?1 loofon_at_[hidden])(?2 "name"<name_at_email.101.pi>")
> #g = (?1 loofon_at_[hidden])(?2 "name"<name_at_email.101.pi>")
> 
> rather than
> #f = loofon_at_[hidden]
> #g = "name"<name_at_email.101.pi>"
> 
> In my case, I need to handle a heap of patterns and using replace_all_regex is convenient.  Any way to use both
> replace_all_regex and format_literal without going into the trouble mentioned above?
> 

The only solution that I can comeup with is to run replace_all_regex
twice. Once for #f pattern and the other time for the #g pattern.

If you use format_literal, it is obvious, that (?1 and (?2 format
specifiers are not evaluated.

Maybe there is another solution for this, but I don't know about it.

Best regards,
Pavol.