C++ Replace part of a string with another string

Question

Is it possible in C++ to replace part of a string with another string. Basically, I would like to do this

QString string("hello $name");
string.replace("$name", "Somename");

but I would like to use the Standard C++ libraries.

Answer 1

std::string has a replace method, is that what you are looking for?

You could try:

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

I haven't tried myself, just read the documentation on find() and replace()

Answer 2

Yes, you can do it, but you have to find the position of the first string with string's find() member, and then replace with it's replace() member.

string s("hello $name");
size_type pos = s.find( "$name" );
if ( pos != string::npos ) {
   s.replace( pos, 5, "somename" );   // 5 = length( $name )
}

If you are planning on using the Standard Library, you should really gt hold of a copy of the book The C++ Standard Library which covers all this stuff very well.

Answer 3

There's a function to find a substring within a string (find), and a function to replace a particular range in a string with another string (replace), so you can combine those to get the effect you want:

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

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

---

In response to a comment, I think replaceAll would probably look something like this:

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