我自己的实现，考虑到字符串只需要调整一次大小，然后替换就可以发生。

template <typename T>
std::basic_string<T> replaceAll(const std::basic_string<T>& s, const T* from, const T* to)
{
    auto length = std::char_traits<T>::length;
    size_t toLen = length(to), fromLen = length(from), delta = toLen - fromLen;
    bool pass = false;
    std::string ns = s;

    size_t newLen = ns.length();

    for (bool estimate : { true, false })
    {
        size_t pos = 0;

        for (; (pos = ns.find(from, pos)) != std::string::npos; pos++)
        {
            if (estimate)
            {
                newLen += delta;
                pos += fromLen;
            }
            else
            {
                ns.replace(pos, fromLen, to);
                pos += delta;
            }
        }

        if (estimate)
            ns.resize(newLen);
    }

    return ns;
}

用法可以是这样的:

std::string dirSuite = replaceAll(replaceAll(relPath.parent_path().u8string(), "\\", "/"), ":", "");

如果所有字符串都是std::string，如果使用sizeof()，你会发现字符截断的奇怪问题，因为它是针对C字符串，而不是c++字符串。修复方法是使用std::string的.size()类方法。

sHaystack.replace(sHaystack.find(sNeedle), sNeedle.size(), sReplace);

这代替了sHaystack内联-不需要做=赋值回来。

使用示例:

std::string sHaystack = "This is %XXX% test.";
std::string sNeedle = "%XXX%";
std::string sReplace = "my special";
sHaystack.replace(sHaystack.find(sNeedle),sNeedle.size(),sReplace);
std::cout << sHaystack << std::endl;

std::string replace(std::string base, const std::string from, const std::string to) {
    std::string SecureCopy = base;

    for (size_t start_pos = SecureCopy.find(from); start_pos != std::string::npos; start_pos = SecureCopy.find(from,start_pos))
    {
        SecureCopy.replace(start_pos, from.length(), to);
    }

    return SecureCopy;
}

wstring myString = L"Hello $$ this is an example. By $$.";
wstring search = L"$$";
wstring replace = L"Tom";
for (int i = myString.find(search); i >= 0; i = myString.find(search))
    myString.replace(i, search.size(), replace);

要返回新字符串，请使用以下命令:

std::string ReplaceString(std::string subject, const std::string& search,
                          const std::string& replace) {
    size_t pos = 0;
    while ((pos = subject.find(search, pos)) != std::string::npos) {
         subject.replace(pos, search.length(), replace);
         pos += replace.length();
    }
    return subject;
}

如果你需要性能，这里有一个修改输入字符串的优化函数，它不会创建字符串的副本:

void ReplaceStringInPlace(std::string& subject, const std::string& search,
                          const std::string& replace) {
    size_t pos = 0;
    while ((pos = subject.find(search, pos)) != std::string::npos) {
         subject.replace(pos, search.length(), replace);
         pos += replace.length();
    }
}

测试:

std::string input = "abc abc def";
std::cout << "Input string: " << input << std::endl;

std::cout << "ReplaceString() return value: " 
          << ReplaceString(input, "bc", "!!") << std::endl;
std::cout << "ReplaceString() input string not modified: " 
          << input << std::endl;

ReplaceStringInPlace(input, "bc", "??");
std::cout << "ReplaceStringInPlace() input string modified: " 
          << input << std::endl;

输出:

Input string: abc abc def
ReplaceString() return value: a!! a!! def
ReplaceString() input string not modified: abc abc def
ReplaceStringInPlace() input string modified: a?? a?? def

有一个函数用于查找字符串中的子字符串(find)，还有一个函数用于用另一个字符串替换字符串中的特定范围(replace)，所以你可以结合它们来得到你想要的效果:

bool replace(std::string& str, const std::string& from, const std::string& to) {
    size_t start_pos = str.find(from);
    if(start_pos == std::string::npos)
        return false;
    str.replace(start_pos, from.length(), to);
    return true;
}

std::string string("hello $name");
replace(string, "$name", "Somename");

作为对注释的回应，我认为replaceAll可能看起来像这样:

void replaceAll(std::string& str, const std::string& from, const std::string& to) {
    if(from.empty())
        return;
    size_t start_pos = 0;
    while((start_pos = str.find(from, start_pos)) != std::string::npos) {
        str.replace(start_pos, from.length(), to);
        start_pos += to.length(); // In case 'to' contains 'from', like replacing 'x' with 'yx'
    }
}