Question

i want to replace a character in the string with a string. can i do it in-place? As the new string has length greater than original string.Question is that can i do with using additional buffer?
for example

void replaceChar(std::string &input, std::string replacementString, char charToReplace)
{
//some code here. No additional buffer
}

void main(){

  std::string input = "I am posting a comment on LinkedIn";
  std::string replacementString = "pppp";
  char charToReplace = 'o';
  replaceChar(input, replacementString, charToReplace);
}

I only want the strategy (algorithm). it would be good if algorithm will be designed keeping some language in mind that will not dynamically increase or decrease the string length once it was initilized like c++

Answer 1

std::string has a replace member, but it works in terms of numerical positions, rather than the previous content of the string. As such, you normally have to combine it with the find member in a loop, something like this:

std::string old("o");

int pos;

while ((pos = x.find(old)) != std::string::npos)
    x.replace(pos, old.length(), "pppp");

Personally, I’d rarely get concerned about how often the string gets resized, but if it’s a major concern, you can use std::count to find the number of occurrences of the old string, multiply by the difference in size between the old and new strings, and use std::string::reserve() to reserve enough space. Note, however, that reserve was added in C++11 — older implementations won’t have it.

Alhough it’s not a concern with the strings you used, this doesn’t work correctly if the replacement string contains an instance of the value being replaced. If you need to deal with that, you’ll need to supply the offset in the string at which to start each search:

int pos = 0;

while ((pos = x.find(old, pos)) != std::string::npos) {
    x.replace(pos, old.length(), rep);
    pos += rep.length();
}

Or, you might prefer a for loop in this case:

    std::string old("o");
    std::string rep("pop");

for (std::size_t pos=0; 
    (pos = x.find(old, pos)) != std::string::npos; 
    pos+=rep.length())
{
    x.replace(pos, old.length(), rep);
}