[c++] How to convert a command-line argument to int?

I need to get an argument and convert it to an int. Here is my code so far:

#include <iostream>


using namespace std;
int main(int argc,int argvx[]) {
    int i=1;
    int answer = 23;
    int temp;

    // decode arguments
    if(argc < 2) {
        printf("You must provide at least one argument\n");
        exit(0);
    }

    // Convert it to an int here

}

As WhirlWind has pointed out, the recommendations to use atoi aren't really very good. atoi has no way to indicate an error, so you get the same return from atoi("0"); as you do from atoi("abc");. The first is clearly meaningful, but the second is a clear error.

He also recommended strtol, which is perfectly fine, if a little bit clumsy. Another possibility would be to use sscanf, something like:

if (1==sscanf(argv[1], "%d", &temp))
    // successful conversion
else
    // couldn't convert input

note that strtol does give slightly more detailed results though -- in particular, if you got an argument like 123abc, the sscanf call would simply say it had converted a number (123), whereas strtol would not only tel you it had converted the number, but also a pointer to the a (i.e., the beginning of the part it could not convert to a number).

Since you're using C++, you could also consider using boost::lexical_cast. This is almost as simple to use as atoi, but also provides (roughly) the same level of detail in reporting errors as strtol. The biggest expense is that it can throw exceptions, so to use it your code has to be exception-safe. If you're writing C++, you should do that anyway, but it kind of forces the issue.

Like that we can do....

int main(int argc, char *argv[]) {

    int a, b, c;
    *// Converting string type to integer type
    // using function "atoi( argument)"* 

    a = atoi(argv[1]);     
    b = atoi(argv[2]);
    c = atoi(argv[3]);

 }

Note that your main arguments are not correct. The standard form should be:

int main(int argc, char *argv[])

or equivalently:

int main(int argc, char **argv)

There are many ways to achieve the conversion. This is one approach:

#include <sstream>

int main(int argc, char *argv[])
{
    if (argc >= 2)
    {
        std::istringstream iss( argv[1] );
        int val;

        if (iss >> val)
        {
            // Conversion successful
        }
    }

    return 0;
}

Take a look at strtol(), if you're using the C standard library.

The approach with istringstream can be improved in order to check that no other characters have been inserted after the expected argument:

#include <sstream>

int main(int argc, char *argv[])
{
    if (argc >= 2)
    {
        std::istringstream iss( argv[1] );
        int val;

        if ((iss >> val) && iss.eof()) // Check eofbit
        {
            // Conversion successful
        }
    }

    return 0;
}

std::stoi from string could also be used.

    #include <string>

    using namespace std;

    int main (int argc, char** argv)
    {
         if (argc >= 2)
         {
             int val = stoi(argv[1]);
             // ...    
         }
         return 0;
    }