C++, Boost regex, replace value function of matched value?

Question

Specifically, I have an array of strings called val, and want to replace all instances of "%{n}%" in the input with val[n]. More generally, I want the replace value to be a function of the match value. This is in C++, so I went with Boost, but if another common regex library matches my needs better let me know.

I found some .NET (C#, VB.NET) solutions, but I don't know if I can use the same approach here (or, if I can, how to do so).

I know there is this ugly solution: have an expression of the form "(%{0}%)|(%{1}%)..." and then have a replace pattern like "(1?" + val[0] + ")(2?" + val[1] ... + ")".

But I'd like to know if what I'm trying to do can be done more elegantly.

Thanks!

Answer 1

I don't beleive boost::regex has an easy way to do this. The most straightfoward way that I can think of would be to do a regex_search using the "(%{[0-9]+}%)" pattern and then iterate over the sub-matches in the returned match_results object. You'll need to build a new string by concatenating the text from between each match (the match_results::position method will be your friend here) with the result of converting sub-matches to the values from your val array.