- In C++11, use
std::stoi
as:std::string s = "10"; int i = std::stoi(s);
Note thatstd::stoi
will throw exception of typestd::invalid_argument
if the conversion cannot be performed, orstd::out_of_range
if the conversion results in overflow(i.e when the string value is too big forint
type). You can usestd::stol
orstd:stoll
though in caseint
seems too small for the input string. - In C++03/98, any of the following can be used:
std::string s = "10"; int i; //approach one std::istringstream(s) >> i; //i is 10 after this //approach two sscanf(s.c_str(), "%d", &i); //i is 10 after this
Note that the above two approaches would fail for input s = "10jh"
. They will return 10 instead of notifying error. So the safe and robust approach is to write your own function that parses the input string, and verify each character to check if it is digit or not, and then work accordingly. Here is one robust implemtation (untested though):
int to_int(char const *s) { if ( s == NULL || *s == '\0' ) throw std::invalid_argument("null or empty string argument"); bool negate = (s[0] == '-'); if ( *s == '+' || *s == '-' ) ++s; if ( *s == '\0') throw std::invalid_argument("sign character only."); int result = 0; while(*s) { if ( *s < '0' || *s > '9' ) throw std::invalid_argument("invalid input string"); result = result * 10 - (*s - '0'); //assume negative number ++s; } return negate ? result : -result; //-result is positive! }
This solution is slightly modified version of my another solution.