Reputation: 6655
I have string like "y.x-name', where y and x are number ranging from 0 to 100. From this string, what would be the best method to extract 'x' into an integer variable in C++.
Upvotes: 0
Views: 5135
Reputation: 45410
You could split the string by .
and convert it to integer type directly. The second number in while loop is the one you want, see sample code:
template<typename T>
T stringToDecimal(const string& s)
{
T t = T();
std::stringstream ss(s);
ss >> t;
return t;
}
int func()
{
string s("100.3-name");
std::vector<int> v;
std::stringstream ss(s);
string line;
while(std::getline(ss, line, '.'))
{
v.push_back(stringToDecimal<int>(line));
}
std::cout << v.back() << std::endl;
}
It will output: 3
Upvotes: 1
Reputation: 61327
Use two calls to unsigned long strtoul( const char *str, char **str_end, int base )
, e.g:
#include <cstdlib>
#include <iostream>
using namespace std;
int main(){
char const * s = "1.99-name";
char *endp;
unsigned long l1 = strtoul(s,&endp,10);
if (endp == s || *endp != '.') {
cerr << "Bad parse" << endl;
return EXIT_FAILURE;
}
s = endp + 1;
unsigned long l2 = strtoul(s,&endp,10);
if (endp == s || *endp != '-') {
cerr << "Bad parse" << endl;
return EXIT_FAILURE;
}
cout << "num 1 = " << l1 << "; num 2 = " << l2 << endl;
return EXIT_FAILURE;
}
Upvotes: 0
Reputation: 81
You can achieve it with boost::lexical_cast, which utilizes streams like in billz' answer: Pseudo code would look like this (indices might be wrong in that example):
std::string yxString = "56.74-name";
size_t xStart = yxString.find(".") + 1;
size_t xLength = yxString.find("-") - xStart;
int x = boost::lexical_cast<int>( yxString + xStart, xLength );
Parsing errors can be handled via exceptions that are thrown by lexical_cast. For more flexible / powerful text matching I suggest boost::regex.
Upvotes: 0
Reputation: 1135
It seem that this thread has a problem similar to you, it might help ;)
Simple string parsing with C++
Upvotes: 0