How can I get argv[] as int?

Question

i have a piece of code like this:

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

Answer 1

/*

    Input from command line using atoi, and strtol 
*/

#include //printf, scanf
#include //atoi, strtol 

//strtol - converts a string to a long int 
//atoi - converts string to an int 

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

    char *p;//used in strtol 
    int i;//used in for loop

    long int longN = strtol( argv[1],&p, 10);
    printf("longN = %ld\n",longN);

    //cast (int) to strtol
    int N = (int) strtol( argv[1],&p, 10);
    printf("N = %d\n",N);

    int atoiN;
    for(i = 0; i < argc; i++)
    {
        //set atoiN equal to the users number in the command line 
        //The C library function int atoi(const char *str) converts the string argument str to an integer (type int).
        atoiN = atoi(argv[i]);
    }

    printf("atoiN = %d\n",atoiN);
    //-----------------------------------------------------//
    //Get string input from command line 
    char * charN;

    for(i = 0; i < argc; i++)
    {
        charN = argv[i];
    }

    printf("charN = %s\n", charN); 

}

Hope this helps. Good luck!