For example, if I have
string x = "dog:cat"; and I want to extract everything after the ":", and return cat. What would be the way to go about doing this?
- 1See the example here: cplusplus.com/reference/string/string/substrDronz– Dronz2015-01-27 05:24:47 +00:00Commented Jan 27, 2015 at 5:24
- This question could have been avoided with a bit of research.sjsam– sjsam2015-01-27 05:41:48 +00:00Commented Jan 27, 2015 at 5:41
- 10Maybe so, but it's top of Google results now.Ben Fulton– Ben Fulton2017-03-31 14:04:59 +00:00Commented Mar 31, 2017 at 14:04
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9 Answers
Try this:
x.substr(x.find(":") + 1); 
5 Comments
Spandyie
This is amazing !! <3
Kyle
This creates a copy, which may not be what you want. In C++17 you can use
std::string_view to avoid copying.
galsh83
As mentioned in some of the other answers, you should handle the edge case of
find returning npos. It is not guaranteed that npos + 1 equals 0 (see stackoverflow.com/questions/26402961/…).
User123
What if I had "cat:dog:parrot:horse" and I want to get just horse? (So the last one :)
Virtuall.Kingg
@rcs How should I do exactly same thing in C programming?
I know it will be super late but I am not able to comment accepted answer. If you are using only a single character in find function use '' instead of "". As Clang-Tidy says The character literal overload is more efficient.
So x.substr(x.find(':') + 1)
Comments
The accepted answer from rcs can be improved. Don't have rep so I can't comment on the answer.
std::string x = "dog:cat"; std::string substr; auto npos = x.find(":"); if (npos != std::string::npos) substr = x.substr(npos + 1); if (!substr.empty()) ; // Found substring; Not performing proper error checking trips up lots of programmers. The string has the sentinel the OP is interested but throws std::out_of_range if pos > size().
basic_string substr( size_type pos = 0, size_type count = npos ) const; 
Comments
#include <iostream> #include <string> int main(){ std::string x = "dog:cat"; //prints cat std::cout << x.substr(x.find(":") + 1) << '\n'; } Here is an implementation wrapped in a function that will work on a delimiter of any length:
#include <iostream> #include <string> std::string get_right_of_delim(std::string const& str, std::string const& delim){ return str.substr(str.find(delim) + delim.size()); } int main(){ //prints cat std::cout << get_right_of_delim("dog::cat","::") << '\n'; } 
1 Comment
sjsam
Thinking of delims of any length is simply smart, so a +1
something like this:
string x = "dog:cat"; int i = x.find_first_of(":"); string cat = x.substr(i+1); Comments
Try this:
string x="dog:cat"; int pos = x.find(":"); string sub = x.substr (pos+1); cout << sub; 
Comments
What you can do is get the position of ':' from your string, then retrieve everything after that position using substring.
size_t pos = x.find(":"); // position of ":" in str
string str3 = str.substr (pos);
Comments
#include <string> #include <iostream> std::string process(std::string const& s) { std::string::size_type pos = s.find(':'); if (pos!= std::string::npos) { return s.substr(pos+1,s.length()); } else { return s; } } int main() { std::string s = process("dog:cat"); std::cout << s; } 
1 Comment
Brahmanand Choudhary
I have executed this code it works ..tutorialspoint.com/compile_cpp_online.php
Try this one..
std::stringstream x("dog:cat"); std::string segment; std::vector<std::string> seglist; while(std::getline(x, segment, ':')) { seglist.push_back(segment); } 