std::string
has a replace
member, but it works in terms of numerical positions, rather than the previous content of the string. As such, you normally have to combine it with the find
member in a loop, something like this:
std::string old("o");
int pos;
while ((pos = x.find(old)) != std::string::npos)
x.replace(pos, old.length(), "pppp");
Personally, I'd rarely get concerned about how often the string gets resized, but if it's a major concern, you can use std::count
to find the number of occurrences of the old
string, multiply by the difference in size between the old and new strings, and use std::string::reserve()
to reserve enough space. Note, however, that reserve
was added in C++11 -- older implementations won't have it.
Edit: though it's not a concern with the strings you used, as @ipc pointed out, this doesn't work correctly if the replacement string contains an instance of the value being replaced. If you need to deal with that, you'll need to supply the offset in the string at which to start each search:
int pos = 0;
while ((pos = x.find(old, pos)) != std::string::npos) {
x.replace(pos, old.length(), rep);
pos += rep.length();
}
Or, you might prefer a for
loop in this case:
std::string old("o");
std::string rep("pop");
for (int pos=0;
(pos = x.find(old, pos)) != std::string::npos;
pos+=rep.length())
{
x.replace(pos, old.length(), rep);
}
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