How to efficiently replace characters in an std::string with iterative values from another std::string?

user_5731890 0
Viewed 302

I have the following strings:

std::string str1 = "1234567890";
std::string str2 = "B-XXXX_XXX_V-XX_X";

I want to loop through str2 and replace every occurrence of X with the subsequent value from str1, resulting in: B-1234_567_V-89_0.

I have a semblance of a solution below, but it's not very efficient (it worked at one point). In brief, I tried to loop through the characters in str2, and if the character equaled 'X', replace that character with an incrementing index from str1:

  int ind = 0;
  std::string pattern_char;
  for (int i = 0; i < str2.size(); i++) {
    pattern_char = str2[i];
    if (pattern_char == "X") {
      str2[i] = str1[x_ind];
      x_ind++;
    }
  }

What is the most efficient way to perform this operation?

stdstring loops c++
4 Answers

Your current implementation treats each individual unit of the string as a std::string rather than as a single char, which introduces some unnecessary overhead. Here's a rewrite that uses chars:

  int x_ind = 0;
  for (int i = 0; i < str2.size(); i++) {
    if (str2[i] == 'X') {    // Don't assign str2[i] to a char; use character literals
      str2[i] = str1[x_ind];
      x_ind++;
    }
  }

You can improve readability by using a range-based for loop, like this:

  int x_ind = 0;
  for (char& ch: str2) {
    if (ch == 'X') {
      ch = str1[x_ind];
      x_ind++;
    }
  }
user_5015570

If by "not very efficient" you want to improve your current code, maybe the only thing to do is to rewrite your loop:

int idx = 0;
for (char& c : str2) {
    if (c == 'X')
      c = str1[idx++];
}

But if you want to write this by only using standard library, you can do the same thing by using std::transform():

#include <algorithm>
#include <iostream>
#include <string>

int main()
{
    std::string str1 = "1234567890";
    std::string str2 = "B-XXXX_XXX_V-XX_X";
    int i = 0;
    std::transform(str2.begin(), str2.end(), str2.begin(),
                   [&str1, &i](const char& c) -> char {
                       return c == 'X' ? str1[i++] : c;
                   });

    std::cout << str2 << std::endl;
}
user_76110280

The only problem I see with your code is that you unnecessary use std::string pattern_char;:

#include <string>
#include <iostream>
#include <stdexcept>

auto replace_with_pattern(std::string& str, char ch_to_replace, const std::string& pattern)
{
    auto pattern_it = pattern.begin();

    for (char& ch : str)
    {
        if (ch == ch_to_replace)
        {
            if (pattern_it == pattern.end())
                throw std::invalid_argument{"ran out of pattern"};

            ch = *pattern_it;
            ++pattern_it;
        }
    }
}

int main()
{
    std::string str1 = "1234567890";
    std::string str2 = "B-XXXX_XXX_V-XX_X";


    replace_with_pattern(str2, 'X', str1);

    std::cout << str2 << std::endl;
}
user_28053050

Here's another solution:

auto i = str1.begin();
for (char& ch: str2) {
  if (ch == 'X') 
    ch = *i++;
user_83728530
Related