什么是有效的方法来取代一个字符的所有出现与另一个字符在std::字符串?


当前回答

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;"}});

其他回答

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, "#", "@");

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

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'
}

如何替换任何字符串与任何字符串仅使用良好的旧C字符串函数?

char original[256]="First Line\nNext Line\n", dest[256]="";
char* replace_this = "\n"; // this is now a single character but could be any string
char* with_this = "\r\n"; // this is 2 characters but could be of any length

/* get the first token */
char* token = strtok(original, replace_this);

/* walk through other tokens */
while (token != NULL) {
    strcat(dest, token);
    strcat(dest, with_this);
    token = strtok(NULL, replace_this);
}

Dest现在应该有我们要找的东西了。