Regex replace names of methods

Question

I'm trying to replace all occurrences of names within a given string. I'm using regex, since a simple substring match won't work in this case and I need to match full words.

My problem is that I can only match words before and after blanks. But for example I cannot replace a string when it's followed by a blank, like:

toReplace()

with: theReplacement()

My regex replace method looks like this:

void replaceWord(std::string &str, const std::string& search, const std::string& replace)
{
    // Regular expression to match words beginning with 'search'
    //    std::regex e ("(\\b("+search+"))([^,. ]*)");
    //    std::regex e ("(\\b("+search+"))\\b)");
    std::regex e("(\\b("+search+"))([^,.()<>{} ]*)");
    str = std::regex_replace(str,e,replace) ;
}

How should the regex look like in order to ignore leading and trailing non-alphanumericals?

Could you please add some test cases that show what the program does incorrectly, and what it should do instead? — cigien
– cigien, Commented Aug 28, 2020 at 16:49
@Eljay I've tried that (see commented code above). It doesn't work for cases like this: ideone.com/iMFKfL — benjist
– benjist, Commented Aug 28, 2020 at 17:47
@benjist • it worked on my machine, where \\b correctly broke only on a word boundary. — Eljay
– Eljay, Commented Aug 28, 2020 at 18:07

Wiktor Stribiżew · Accepted Answer · 2020-08-28 19:14:38Z

2

You need to

Escape all special characters in the regex pattern with std::regex_replace(search, std::regex(R"([.^$|{}()[\]*+?/\\])"), std::string(R"(\$&)"))
Escape all special chars in the replacement pattern with std::regex_replace(replace, std::regex("[$]"), std::string("$$$$")) (that is in case you replace with literal $1 text, $ can be set with $$, so to replace with a double $, we need $$$$ in the replacement here)
Wrap your search pattern with unambiguous word boundaries, i.e. "(\\W|^)("+search+")(?!\\w)
When you replace, add $1 at the start of the replacement pattern to keep the whitespace (if it is matched and captured into the first group with the (\W|^) pattern).

See C++ sample code:

std::string replaceWord(std::string &str, std::string& search, std::string& replace)
{
    // Escape the literal regex pattern
    search = std::regex_replace(search, std::regex(R"([.^$|{}()[\]*+?/\\])"), std::string(R"(\$&)"));
    // Escape the literal replacement pattern
    replace = std::regex_replace(replace, std::regex("[$]"), std::string("$$$$"));
    std::regex e("(\\W|^)("+search+")(?!\\w)");
    return std::regex_replace(str, e, std::string("$1") + replace);
}

Then,

std::string text("String toReplace()");
std::string s("toReplace()");
std::string r("theReplacement()");
std::cout << replaceWord(text, s, r);   
// => String theReplacement()

edited Aug 28, 2020 at 19:14

answered Aug 28, 2020 at 17:08

Wiktor Stribiżew

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2 Comments

benjist Over a year ago

Thanks for that. Though it does not work when trying to match a word (without brackets), and when there are non-alpha characters before it. I've made an example: ideone.com/iMFKfL

Wiktor Stribiżew Over a year ago

@benjist Then use unambiguous word boundaries rather than whitespace boundaries. See this C++ demo.

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