c - How can I get argv[] as int?

Question

i have a piece of code like this:

int main (int argc, char *argv[]) 
{
   printf("%d",(int)argv[1]);
   printf("%s",(int)argv[1]);
}

and in shell i do this:

./test 7

but the first printf result is not 7, how can I get argv[] as a int? many thanks

Answer 1

argv[1] is a pointer to a string.

You can print the string it points to using printf("%s ", argv[1]);

To get an integer from a string you have first to convert it. Use strtol to convert a string to an int.

#include <stdio.h>
#include <errno.h>   // for errno
#include <limits.h>  // for INT_MAX, INT_MIN
#include <stdlib.h>  // for strtol


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

    errno = 0;
    long conv = strtol(argv[1], &p, 10);

    // Check for errors: e.g., the string does not represent an integer
    // or the integer is larger than int
    if (errno != 0 || *p != '' || conv > INT_MAX || conv < INT_MIN) {
        // Put here the handling of the error, like exiting the program with
        // an error message
    } else {
        // No error
        num = conv;
        printf("%d
", num);
    }
}