为了完整起见，下面是如何使用std::regex来实现它。

#include <regex>
#include <string>

int main()
{
    const std::string s = "example string";
    const std::string r = std::regex_replace(s, std::regex("x"), "y");
}

老派:-)

std::string str = "H:/recursos/audio/youtube/libre/falta/"; 

for (int i = 0; i < str.size(); i++) {
    if (str[i] == '/') {
        str[i] = '\\';
    }
}

std::cout << str;

结果:

点:youtube \ resources \音响\ \‘\ \缺失

我想我也会加入促进方案:

#include <boost/algorithm/string/replace.hpp>

// in place
std::string in_place = "blah#blah";
boost::replace_all(in_place, "#", "@");

// copy
const std::string input = "blah#blah";
std::string output = boost::replace_all_copy(input, "#", "@");

#include <iostream>
#include <string>
using namespace std;
// Replace function..
string replace(string word, string target, string replacement){
    int len, loop=0;
    string nword="", let;
    len=word.length();
    len--;
    while(loop<=len){
        let=word.substr(loop, 1);
        if(let==target){
            nword=nword+replacement;
        }else{
            nword=nword+let;
        }
        loop++;
    }
    return nword;

}
//Main..
int main() {
  string word;
  cout<<"Enter Word: ";
  cin>>word;
  cout<<replace(word, "x", "y")<<endl;
  return 0;
}

这是我滚动的一个解决方案，在最大的DRI精神。 它将在sHaystack中搜索sNeedle并将其替换为sReplace， nTimes如果不为0，否则所有的sNeedle发生。 它不会在替换的文本中再次搜索。

std::string str_replace(
    std::string sHaystack, std::string sNeedle, std::string sReplace, 
    size_t nTimes=0)
{
    size_t found = 0, pos = 0, c = 0;
    size_t len = sNeedle.size();
    size_t replen = sReplace.size();
    std::string input(sHaystack);

    do {
        found = input.find(sNeedle, pos);
        if (found == std::string::npos) {
            break;
        }
        input.replace(found, len, sReplace);
        pos = found + replen;
        ++c;
    } while(!nTimes || c < nTimes);

    return input;
}

为了完整起见，下面是如何使用std::regex来实现它。

#include <regex>
#include <string>

int main()
{
    const std::string s = "example string";
    const std::string r = std::regex_replace(s, std::regex("x"), "y");
}