How do I use atoi function with strings in C++

Question

This is a basic question. I use C++ but not C++11. Now, I want to convert a string to an integer. I have declared like this:

string s;

int i = atoi(s);

However, this shows an error that such a conversion is not possible. I checked out the internet and I found that C++11 has stoi() but I want to use atoi itself. How can I do it? Thanks!

Answer 1

// atoi_string (cX) 2014 adolfo.dimare@gmail.com
// http://stackoverflow.com/questions/27640333/

#include <string>

/// Convert 'str' to its integer value.
/// - Skips leading white space.
int atoi( const std::string& str ) {
    std::string::const_iterator it;
    it = str.begin();
    while ( isspace(*it)) { ++it; } // skip white space
    bool isPositive = true;
    if ( *it=='-' ) {
        isPositive = false;
        ++it;
    }
    int val = 0;
    while ( isdigit(*it) ) {
        val = val * 10 + (*it)-'0';
    }
    return ( isPositive ? val : -val );
}

#include <cassert> // assert()
#include <climits> // INT_MIN && INT_MAX
#include <cstdlib> // itoa()

 int main() {
     char mem[ 1+sizeof(int) ];
     std::string str;
     for ( int i=INT_MIN; true; ++i ) {
        itoa( i, mem, 10 );
        str = mem;
        assert( i==atoi(str) ); // never stops
     }
 }

Answer 2

There are also strtol(), and strtoul() family of functions, you might find them useful for different base etc

Answer 3

Convert it into a C string and you're done

string s;

int i = atoi(s.c_str());

Answer 4

Use instead

int i = atoi( s.c_str() );