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

argv[1] is a pointer to a string.

You can print the string it points to using printf("%s\n", 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    // for errno
#include   // for INT_MAX
#include   // for strtol
 
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 != '\0' || conv > INT_MAX) {
    // Put here the handling of the error, like exiting the program with
    // an error message
} else {
    // No error
    num = conv;    
    printf("%d\n", num);
}