如何用标准c++库替换字符串的一部分?
QString s("hello $name"); // Example using Qt.
s.replace("$name", "Somename");
如何用标准c++库替换字符串的一部分?
QString s("hello $name"); // Example using Qt.
s.replace("$name", "Somename");
当前回答
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;
}
其他回答
那么加速方案呢:
boost::replace_all(value, "token1", "token2");
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(), "\\", "/"), ":", "");
这个用起来可能更好
void replace(string& input, const string& from, const string& to)
{
auto pos = 0;
while(true)
{
size_t startPosition = input.find(from, pos);
if(startPosition == string::npos)
return;
input.replace(startPosition, from.length(), to);
pos += to.length();
}
}
使用std:: string::替换:
s.replace(s.find("$name"), sizeof("$name") - 1, "Somename");
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'
}
}