我想我会使用std::replace_if()

一个简单的字符替换程序(OP要求的)可以使用标准库函数编写。

就地版本:

#include <string>
#include <algorithm>

void replace_char(std::string& in,
                  std::string::value_type srch,
                  std::string::value_type repl)
{
    std::replace_if(std::begin(in), std::end(in),
                    [&srch](std::string::value_type v) { return v==srch; },
                    repl);
    return;
}

如果输入是const字符串，则返回一个副本的重载:

std::string replace_char(std::string const& in,
                         std::string::value_type srch,
                         std::string::value_type repl)
{
    std::string result{ in };
    replace_char(result, srch, repl);
    return result;
}

Std::string不包含这样的函数，但你可以使用独立的替换函数从算法头。

#include <algorithm>
#include <string>

void some_func() {
  std::string s = "example string";
  std::replace( s.begin(), s.end(), 'x', 'y'); // replace all 'x' to 'y'
}

Abseil StrReplaceAll怎么样?在头文件中:

// This file defines `absl::StrReplaceAll()`, a general-purpose string
// replacement function designed for large, arbitrary text substitutions,
// especially on strings which you are receiving from some other system for
// further processing (e.g. processing regular expressions, escaping HTML
// entities, etc.). `StrReplaceAll` is designed to be efficient even when only
// one substitution is being performed, or when substitution is rare.
//
// If the string being modified is known at compile-time, and the substitutions
// vary, `absl::Substitute()` may be a better choice.
//
// Example:
//
// std::string html_escaped = absl::StrReplaceAll(user_input, {
//                                                {"&", "&"},
//                                                {"<", "&lt;"},
//                                                {">", "&gt;"},
//                                                {"\"", "&quot;"},
//                                                {"'", "&#39;"}});

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;
}

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

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");
}