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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.

2016/07/13
1
190
7/13/2016 11:40:07 AM

Accepted Answer

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 = 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");

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) {
    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'
    }
}
2013/08/28
293
8/28/2013 5:37:34 PM


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

You could try:

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

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

2019/05/12

To have the new string returned use this:

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;
}

If you need performance, here is an optimized function that modifies the input string, it does not create a copy of the string:

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();
    }
}

Tests:

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;

Output:

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
2013/02/26

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 get hold of a copy of the book The C++ Standard Library which covers all this stuff very well.

2012/05/09

I use generally this:

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;
}

It repeatedly calls std::string::find() to locate other occurrences of the searched for string until std::string::find() doesn't find anything. Because std::string::find() returns the position of the match we don't have the problem of invalidating iterators.

2015/06/20

This sounds like an option

string.replace(string.find("%s"), string("%s").size(), "Something");

You could wrap this in a function but this one-line solution sounds acceptable. The problem is that this will change the first occurence only, you might want to loop over it, but it also allows you to insert several variables into this string with the same token (%s)

2015/04/04

Source: https://stackoverflow.com/questions/3418231
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