Question

I have this C program that has two functions to solve this Blackjack Solution. But i need to use only function (not including main of course).

Is it possible to get to just one function outside of Main? #include<stdio.h> int isvalid(char card1, char card2); int sum(char card1, char card2); int sum(char card1, char card2) { int result=0; if(card1=='A' && card2=='A') return 12; if(card1<='9' && card1>='2') result+=card1-'0'; if(card1=='T' || card1 == 'J' || card1=='Q' || card1=='K') result+=10; if(card1=='A') result+=11; if(card2<='9' && card2>='2') result+=card2-'0'; if(card2=='T' || card2 == 'J' || card2=='Q' || card2=='K') result+=10; if(card2=='A') result+=11; return result; } int isvalid(char card1, char card2) { int i,cnt=0; char a[]="23456789ATJQK"; for(i=0;a[i];i++) { if(card1==a[i]) cnt++; if(card2==a[i]) cnt++; } if(cnt==2) return 1; return 0; } int main() { char c1, c2; scanf("%c %c",&c1,&c2); if(isvalid(c1,c2)) printf("%d\n",sum(c1,c2)); else printf("Invalid input"); return 0; }

Answer 1

Yes, it can be solved using one function.

The only function I have used is the 'invalid' function.

I have not modified your function but instead I have used a variable to trace the 'sum' function.

My isvalid function is:

int isvalid(char card1, char card2)
{
   int i,cnt = 0, result = 0;
   char a[]="23456789ATJQK";
   for(i=0;a[i];i++)
   {
       if(card1==a[i])
           cnt++;
       if(card2==a[i])
           cnt++;
   }
   if(cnt==2)
   {
       if(card1=='A' && card2=='A')
           return 12;
       if(card1<='9' && card1>='2')
           result+=card1-'0';
       if(card1=='T' || card1 == 'J' || card1=='Q' || card1=='K')
           result+=10;
       if(card1=='A')
           result+=11;
       if(card2<='9' && card2>='2')
           result+=card2-'0';
       if(card2=='T' || card2 == 'J' || card2=='Q' || card2=='K')
           result+=10;
       if(card2=='A')
           result+=11;
       if(result == 0) // if result is 0 but count is not zero, valid case but result 0,
           return -1; // return value is made different than invalid case, in invalid
                   // case the function returns 0 but when there is valid case but the
                   // result is 0 then that value has to be different than invalid case
                   // so -1 is returned in this case
       return result;
   }
   return cnt; // return value of invalid case is 0
}

The overall program is:

#include<stdio.h>

int isvalid(char card1, char card2)
{
   int i,cnt = 0, result = 0;
   char a[]="23456789ATJQK";
   for(i=0;a[i];i++)
   {
       if(card1==a[i])
           cnt++;
       if(card2==a[i])
           cnt++;
   }
   if(cnt==2)
   {
       if(card1=='A' && card2=='A')
           return 12;
       if(card1<='9' && card1>='2')
           result+=card1-'0';
       if(card1=='T' || card1 == 'J' || card1=='Q' || card1=='K')
           result+=10;
       if(card1=='A')
           result+=11;
       if(card2<='9' && card2>='2')
           result+=card2-'0';
       if(card2=='T' || card2 == 'J' || card2=='Q' || card2=='K')
           result+=10;
       if(card2=='A')
           result+=11;
       if(result == 0) // if result is 0 but count is not zero, valid case but result 0,
           return -1; // return value is made different than invalid case, in invalid
                   // case the function returns 0 but when there is valid case but the
                   // result is 0 then that value has to be different than invalid case
                   // so -1 is returned in this case
       return result;
   }
   return cnt; // return value of invalid case is 0
}
int main()
{
   char c1, c2;
   scanf("%c %c",&c1,&c2);
   int a = isvalid(c1,c2);
   if(a)
   {
       if(a != -1)
           printf("%d\n",a);
       else
           printf("0\n");
   }
   else
       printf("Invalid input");
   return 0;
}

The sample of the program is:

The sample output of the code is:

The program can be shortened further but I have tried to keep it simple for your better understanding.