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C Programming. Fill in ...

This program will be called with one command line argument that
contains a string followed by an asterisk and an integer. Print
out the string as many time as indicated by the integer. For
example, when called as prog Hi*3, you print HiHiHi.

Hint: Look for the '*' starting from the back of the string.
len = strlen(arg) gives you the string length.
When you have found the '*', then arg + len + 1 is a pointer
to the integer that you can pass to atoi as in the preceding
exercise. Replace the '*' with a '\0' so you can print the
part that precedes it.
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char* argv[])
{
   int count = 1;
   char* arg;
   int len;
   if (argc < 2) return -1;
   arg = argv[1];
   len = strlen(arg) - 1;
   ...
   for (int i = 0; i < count; i++)
      printf("%s", arg);
   printf("\n");
   return 0;
}

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char* argv[])
{
   char* arg; int count = 0;
   const char searchStr[2] = "*";
   char *tokens; int inc=0, j;
   char *strArray[3];

   if (argc < 2) return 0;

   arg = argv[1];
   tokens = strtok(arg, searchStr); //-- split string with delimerter of '*' and create new array of string and int
   while( tokens != NULL )
   {
        strArray[inc] = tokens;
       tokens = strtok(NULL, searchStr);
       inc++;
   }

   /*Find string array length*/
   count = sizeof(strArray)/sizeof(strArray[0])-1;

   if (count > 0 && strArray[0] != NULL && atoi(strArray[count-1]) > 0)
   {
       for (j=0; j<atoi(strArray[1]); j++)
       {
           printf("%s", strArray[0]);
       }
   }

   printf("\n");

   return 0;
}

Sample Output:
./a.out Hai*10

HaiHaiHaiHaiHaiHaiHaiHaiHaiHai