How to look for std :: string for substring in C ++?

Question

I am trying to parse a simple string in C ++. I know that a line contains colon text followed by a space immediately followed by a number. I would like to extract only part of the line number. I can't just indicate a space (using sstream and <<), because the text before the colon may or may not contain spaces.

Some lines may be:

Total disk space: 9852465

Free disk space: 6243863

Sectors: 4095

I would like to use the standard library, but if you have another solution, you can post this too, as others with the same question may want to see different solutions.

Answer 1

std::string strInput = "Total disk space: 9852465"; std::string strNumber = "0"; size_t iIndex = strInput.rfind(": "); if(iIndex != std::string::npos && strInput.length() >= 2) { strNumber = strInput.substr(iIndex + 2, strInput.length() - iIndex - 2) } 

Answer 2

For completeness, here's a simple solution in C:

 int value; if(sscanf(mystring.c_str(), "%*[^:]:%d", &value) == 1) // parsing succeeded else // parsing failed 

Explanation: %*[^:] says that as many characters as possible that are not colons are read, but * suppresses the assignment. Then an integer is read after the colon and any intermediate space.

Answer 3

I can't just indicate a space (using sstream and <<), because the text before the colon may or may not contain spaces.

Right, but you can use std::getline :

 string not_number; int number; if (not (getline(cin, not_number, ':') and cin >> number)) { cerr << "No number found." << endl; } 

Answer 4

Like Konrads, but using istream::ignore :

 int number; std::streamsize max = std::numeric_limits<std::streamsize>::max(); if (!(std::cin.ignore(max, ':') >> number)) { std::cerr << "No number found." << std::endl; } else { std::cout << "Number found: " << number << std::endl; } 

Answer 5

I am surprised that no one mentioned regular expressions. They were added as part of TR1 and are included in Boost . Here's a solution using regex's

 typedef std::tr1::match_results<std::string::const_iterator> Results; std::tr1::regex re(":[[:space:]]+([[:digit:]]+)", std::tr1::regex::extended); std::string str("Sectors: 4095"); Results res; if (std::tr1::regex_search(str, res, re)) { std::cout << "Number found: " << res[1] << std::endl; } else { std::cerr << "No number found." << std::endl; } 

It seems like a lot more work, but you will get more from him IMHO.

Answer 6

 const std::string pattern(": "); std::string s("Sectors: 4095"); size_t num_start = s.find(pattern) + pattern.size();