那么加速方案呢:

boost::replace_all(value, "token1", "token2");

我通常用这个:

std::string& replace(std::string& s, const std::string& from, const std::string& to)
{
    if(!from.empty())
        for(size_t pos = 0; (pos = s.find(from, pos)) != std::string::npos; pos += to.size())
            s.replace(pos, from.size(), to);
    return s;
}

它反复调用std::string::find()来定位搜索字符串的其他出现，直到std::string::find()没有找到任何东西。因为std::string::find()返回匹配的位置，所以我们不存在使迭代器失效的问题。

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);

使用std:: string::替换:

s.replace(s.find("$name"), sizeof("$name") - 1, "Somename");

那么加速方案呢:

boost::replace_all(value, "token1", "token2");