Is there a more efficient way to create a counter for a quiz game using C?

Question

I'm trying to develop a quiz type math game in which the user has to solve on 1-5 questions. I want to add a counter that will display all the correct and incorrect answers at the end of the quiz, so far I have been doing it using if else statement, but I feel like there's should be a more efficient way of doing it. My code:

    if ( questions == 2 )
    { 
        printf (" What is 2 + 2\n");
        scanf("%d",&A1);
            if( A1 == 4 )       
            {
                    correct ++;
                }
                else
                {
                    incorrect ++;
            }

        printf (" What is 5 + 2\n");
        scanf("%d",&A2);
            if( A2 == 7 )       
            {
                correct ++;
            }
            else
            {
                incorrect ++;
            }

    }

Here's the code i have the same thing written 5 times for each option that the user can pick. All help is greatly appreciated, thanks in advance!

Answer 1

You could use the total of the questions and substract the correct answers to get the number of incorrect answers. Here is a sample:

#include <stdio.h>
#include <stdlib.h>
int main()
{
   char arQuestions[5][20] = { "2 + 2 =",
                               "2 * 2 =",
                               "2 - 2 =",
                               "2 / 2 =",
                               "2 ^ 2 ="};
   int answers[5] = {4,4,0,1,4};
   int i = 0;
   int answer  = 0;
   int correct = 0;

   for(;i<5;++i)
   {
       printf("%s ", arQuestions[i]);
       if( 1 == scanf("%d", &answer))
         if(answer == answers[i])
            correct++;
       printf("correct<%d> incorrect<%d>\n", correct, (i+1)-correct);
   }
   return(0);
}

Answer 2

Switch statements can be used when you need to check one variable against multiple values (as you do with your if ( questions == 0...5 ). In order not to rewrite the same questions over and over, we can use the way the switch statement naturally overruns into the next case to "cascade" depending on the number of questions we want.

Finally, as some other pointed out, we don't need to separate variables to track correct or incorrect answers; just track correct answers, and at the end the incorrect answers would equal questions - correct. Putting that all together, we have something that looks like this:

int questions;
printf("Number of questions?\n");
scanf("%d",&questions);

int correct = 0;
int answer = 0;

switch(questions) {
    case 5:
        printf("What is 2 + 2?\n");
        scanf("%d",&answer);
        if(answer == 4)       
        {
            correct++;
        }
        //note - no break at end of each case
    case 4:
        //ask another question
    case 3:
        //ask another question
    case 2:
        //ask another question
    case 1:
        //ask another question
    default:
        break;
}
printf("Results:\nCorrect = %d\nIncorrect = %d", correct, questions - correct);

Answer 3

There is another option that might be a little more efficient

you can make correct and incorrect global variables or create a pointer to them, and create another function to check if the entered answer is correct and update correct or incorrect accordingly:

// initiate global variable's
int correct = 0, incorrect = 0;

void checkanswer(int var1, int var2, int useranswer)
{
    if(useranswer == var1 + var2)
        correct++;
    else
        incorrect++;
}

int main(void)
{
...
    printf (" What is 2 + 2\n");
    scanf("%d", &A1);
    checkanswer(2, 2, A1);
//next question
}

This way, instead of repeating yourself, you use the function you wrote.

A few other things:

• Try to find an alternative to scanf, it is a dangerous function that make your code vurnable. See this and or search for more answers cause this topic has a lot of answered questions online.
• I wrote a math game as well, if you want some another example for something similar to what your'e writing. In my game you need to get a score of 10 and the questions are random in any game. See here if your'e interested.

Hope I helped! Feel free to ask me about my game if you have any questions or anything else :)

Answer 4

Here's a more generic version which actually calculates the equation and checks it against the user's answer.

#include <stdio.h>

typedef enum
{
  ADD,
  SUB,
  MUL,
  DIV,
} operation_t;

typedef struct
{
  int op1;
  int op2;
  operation_t op;
} equation_t;

char operation_to_char (operation_t op)
{
  const char CH_OP[] = {'+', '-', '*', '/'};
  return CH_OP[op];
}

int solve (const equation_t* eq)
{
  switch(eq->op)
  {
    case ADD: return eq->op1 + eq->op2;
    case SUB: return eq->op1 - eq->op2;
    case MUL: return eq->op1 * eq->op2;
    case DIV: return eq->op1 / eq->op2;
  }
  return 0; // to silence compiler warning, should never happen
}

int main (void)
{
  const equation_t EQUATIONS[] = 
  {
    {1, 1, ADD},
    {2, 2, ADD},
    {1, 1, SUB},
    {2, 2, MUL},
    {9, 3, DIV},
  };
  const size_t EQUATIONS_N = sizeof(EQUATIONS)/sizeof(*EQUATIONS);

  for(size_t i=0; i<EQUATIONS_N; i++)
  {
    printf("What is %d %c %d? ", 
             EQUATIONS[i].op1,  
             operation_to_char(EQUATIONS[i].op),
             EQUATIONS[i].op2);

    int answer;
    scanf("%d", &answer);
    getchar(); // discard line feed

    int solution = solve(&EQUATIONS[i]);
    if(answer == solution)
    {
      puts("Correct");
    }
    else
    {
      printf("Incorrect, the solution is %d\n", solution);
    }
  }
}

Please note that this code has no error handling of user input.