C program help

C program help

[spiros85]

Hi guys of dear forum!I am a new programmer in C programming language and i
have a question for you about the following C code:
#include <stdio.h>
#include <stdlib.h>

int main()
{
    int x;
    int answer;
    int right;
    int wrong;
    printf("This is a program of sum\n");
    right = 0; /*Initialization of right value*/
    wrong = 0; /*Initialization of wrong value*/
    for(x = 1; x <= 10; x++)[QUOTE][/QUOTE]
    {
          printf("What is %d + %d: ", x, x);
          scanf("%d", &answer);
          printf("The sum is %d\n", answer);
          printf("\a");
          if(answer == x + x)
          {
             printf("%d is RIGHT answer\n", answer);
             right++; /*Saves and increases the amount of right answers*/
          }
          else
          {
             printf("%d is WRONG answer\n", answer);
             b[10] = answer;
             wrong++; /*Saves and increases the amount of wrong answers*/
             printf("The correct answer is %d\n", x + x);
          }
    }
    printf("The right answers are %d and the wrong are %d\n", right, wrong);
    fflush(stdin);
    getchar();
}
 
First, I use DEV C++ compiler!This program asks from user to find the sum of
1+1,2+2 until for loop reaches 10.As you can observe with right++ and
wrong++ i can define and print to the screen the amount of right or wrong
numbers!
My question is how could I define the number of correct and wrong numbers in
collaboration with their amount?For example:
The right answers are 8 and wrong are 2
The right answers are 2 4 6 8 10 12 14 16
The wrong answers are 22 44
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Re: C program help

[Tim Walberg]

Not tested, but here's a simple (not very efficient or elegant, but
reasonable) concept. It's basically keeping a history, and postprocessing
the history to generate the lists. It might be better to keep separate
right/wrong lists as you go, but it's not as simple...


#include <stdio.h>
#include <stdlib.h>
#define NPROBLEMS 10

int main()
{ int x[NPROBLEMS], y[NPROBLEMS], right_answer[NPROBLEMS], given_answer[NPROBLEMS];
  int nright, nwrong;
  int i;

  for (i = 0; i < NPROBLEMS; ++i)
  { x[i] = i;
    y[i] = i;
    right_answer[i] = x[i] + y[i];
    printf("What is %d + %d? ", x[i], y[i]);
    scanf("%d", &(given_answer[i]));

    if (given_answer[i] == right_answer[i])
    { printf("%d is RIGHT answer\n", given_answer[i]);
      ++nright;
    }
    else
    { printf("%d is WRONG answer; the RIGHT answer is %d\n", given_answer[i], right_answer[i]);
      ++nwrong;
    }
  }

  printf("You got %d right and %d wrong\n", nright, nwrong);

  printf("Right answers:");
  for (i = 0; i < NPROBLEMS; ++i)
    if (right_answer[i] == given_answer[i])
      printf(" %d", given_answer[i]);
  printf("\n");

  printf("Wrong answers:");
  for (i = 0; i < NPROBLEMS; ++i)
    if (right_answer[i] != given_answer[i])
      printf(" %d", given_answer[i]);
  printf("\n");

  return 0;
}

  
On 09/19/2009 08:13 -0700, spiros85 wrote:
>>	
>>	Hi guys of dear forum!I am a new programmer in C programming language and i
>>	have a question for you about the following C code:
>>	#include <stdio.h>
>>	#include <stdlib.h>
>>	
>>	int main()
>>	{
>>	    int x;
>>	    int answer;
>>	    int right;
>>	    int wrong;
>>	    printf("This is a program of sum\n");
>>	    right = 0; /*Initialization of right value*/
>>	    wrong = 0; /*Initialization of wrong value*/
>>	    for(x = 1; x <= 10; x++)[QUOTE][/QUOTE]
>>	    {
>>	          printf("What is %d + %d: ", x, x);
>>	          scanf("%d", &answer);
>>	          printf("The sum is %d\n", answer);
>>	          printf("\a");
>>	          if(answer == x + x)
>>	          {
>>	             printf("%d is RIGHT answer\n", answer);
>>	             right++; /*Saves and increases the amount of right answers*/
>>	          }
>>	          else
>>	          {
>>	             printf("%d is WRONG answer\n", answer);
>>	             b[10] = answer;
>>	             wrong++; /*Saves and increases the amount of wrong answers*/
>>	             printf("The correct answer is %d\n", x + x);
>>	          }
>>	    }
>>	    printf("The right answers are %d and the wrong are %d\n", right, wrong);
>>	    fflush(stdin);
>>	    getchar();
>>	}
>>	 
>>	First, I use DEV C++ compiler!This program asks from user to find the sum of
>>	1+1,2+2 until for loop reaches 10.As you can observe with right++ and
>>	wrong++ i can define and print to the screen the amount of right or wrong
>>	numbers!
>>	My question is how could I define the number of correct and wrong numbers in
>>	collaboration with their amount?For example:
>>	The right answers are 8 and wrong are 2
>>	The right answers are 2 4 6 8 10 12 14 16
>>	The wrong answers are 22 44
>>	-- 
>>	View this message in context: http://www.nabble.com/C-program-help-tp25522429p25522429.html
>>	Sent from the linux-c-programming mailing list archive at Nabble.com.
>>	
>>	--
>>	To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
>>	the body of a message to majordomo@vger.kernel.org
>>	More majordomo info at  http://vger.kernel.org/majordomo-info.html
End of included message



-- 
twalberg@comcast.net

Re: C program help

[Manish Katiyar]

On Sat, Sep 19, 2009 at 8:43 PM, spiros85 <spirosfr_1985@hotmail.com> wrote:
>
> Hi guys of dear forum!I am a new programmer in C programming language and i
> have a question for you about the following C code:
> #include <stdio.h>
> #include <stdlib.h>
>
> int main()
> {
>    int x;
>    int answer;
>    int right;
>    int wrong;
>    printf("This is a program of sum\n");
>    right = 0; /*Initialization of right value*/
>    wrong = 0; /*Initialization of wrong value*/
>    for(x = 1; x <= 10; x++)[QUOTE][/QUOTE]
>    {
>          printf("What is %d + %d: ", x, x);
>          scanf("%d", &answer);
>          printf("The sum is %d\n", answer);
>          printf("\a");
>          if(answer == x + x)
>          {
>             printf("%d is RIGHT answer\n", answer);
>             right++; /*Saves and increases the amount of right answers*/
>          }
>          else
>          {
>             printf("%d is WRONG answer\n", answer);
>             b[10] = answer;
>             wrong++; /*Saves and increases the amount of wrong answers*/
>             printf("The correct answer is %d\n", x + x);
>          }
>    }
>    printf("The right answers are %d and the wrong are %d\n", right, wrong);
>    fflush(stdin);
>    getchar();
> }
>
> First, I use DEV C++ compiler!This program asks from user to find the sum of
> 1+1,2+2 until for loop reaches 10.As you can observe with right++ and
> wrong++ i can define and print to the screen the amount of right or wrong
> numbers!
> My question is how could I define the number of correct and wrong numbers in
> collaboration with their amount?For example:
> The right answers are 8 and wrong are 2
> The right answers are 2 4 6 8 10 12 14 16
> The wrong answers are 22 44

#include<stdio.h>
#define N 10
int main() {
	int answer;
	int i;
	int result[N][2];
	int right = 0, wrong = 0;

	for (i=1;i<=N;i++) {
		printf("What is %i + %i ? :",i,i);
		scanf("%d",&answer);
		if (i+i == answer) {
			result[i-1][0] = answer;
			result[i-1][1] = 1;
			right++;
		} else {
			result[i-1][0] = answer;
			result[i-1][1] = 0;
			wrong++;
		}
	}

	printf ("The right answers are %d and wrong answers are %d\n",right, wrong);
	printf("The right answers are : ");
	for (i=0;i<N;i++) {
		if (result[i][1]) {
			printf("%d ", result[i][0]);
		}
	}
	printf("\n");
	printf("The wrong answers are : ");
	for (i=0;i<N;i++) {
		if (!result[i][1]) {
			printf("%d ", result[i][0]);
		}
	}
	printf("\n");
	return 0;
}



> --
> View this message in context: http://www.nabble.com/C-program-help-tp25522429p25522429.html
> Sent from the linux-c-programming mailing list archive at Nabble.com.
>
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
> More majordomo info at  http://vger.kernel.org/majordomo-info.html
>



-- 
Thanks -
Manish
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Re: C program help

[shiva kumar]

here is solution
 
#include <stdio.h>
#include <stdlib.h>

int main()
{
    int x;
    int answer;
    int right;
    int wrong;
    int righta[100];
    int wronga[100];
    int i;

    printf("This is a program of sum\n");
    right = 0; /*Initialization of right value*/
    wrong = 0; /*Initialization of wrong value*/
    for(x = 1; x <= 10; x++)
    {
          printf("What is %d + %d: ", x, x);
          scanf("%d", &answer);
          printf("The sum is %d\n", answer);
          printf("\a");
          if(answer == x + x)
          {
             printf("%d is RIGHT answer\n", answer);
             righta[right]=answer;
             right++; /*Saves and increases the amount of right answers*/
          }
          else
          {
             printf("%d is WRONG answer\n", answer);
             wronga[wrong] = answer;
             wrong++; /*Saves and increases th      }
    }
    righta[right]=0;
    wronga[wrong]=0;
    printf("The right answers are %d and the wrong are %d\n", right, wrong);
    printf("The right answers are ");
    for(i=0;righta[i];i++){
        printf("%d\t",righta[i]);
    }
    printf("\nThe wrong answers are ");
    for(i=0;wronga[i];i++){
        printf("%d\t",wronga[i]);
    }
    fflush(stdin);
    getchar();
}



e amount of wrong answers*/
             printf("The correct answer is %d\n", x + x);
                                                                                              





spiros85 wrote:
> 
> Hi guys of dear forum!I am a new programmer in C programming language and
> i have a question for you about the following C code:
> #include <stdio.h>
> #include <stdlib.h>
> 
> int main()
> {
>     int x;
>     int answer;
>     int right;
>     int wrong;
>     printf("This is a program of sum\n");
>     right = 0; /*Initialization of right value*/
>     wrong = 0; /*Initialization of wrong value*/
>     for(x = 1; x <= 10; x++)[QUOTE][/QUOTE]
>     {
>           printf("What is %d + %d: ", x, x);
>           scanf("%d", &answer);
>           printf("The sum is %d\n", answer);
>           printf("\a");
>           if(answer == x + x)
>           {
>              printf("%d is RIGHT answer\n", answer);
>              right++; /*Saves and increases the amount of right answers*/
>           }
>           else
>           {
>              printf("%d is WRONG answer\n", answer);
>              b[10] = answer;
>              wrong++; /*Saves and increases the amount of wrong answers*/
>              printf("The correct answer is %d\n", x + x);
>           }
>     }
>     printf("The right answers are %d and the wrong are %d\n", right,
> wrong);
>     fflush(stdin);
>     getchar();
> }
>  
> First, I use DEV C++ compiler!This program asks from user to find the sum
> of 1+1,2+2 until for loop reaches 10.As you can observe with right++ and
> wrong++ i can define and print to the screen the amount of right or wrong
> numbers!
> My question is how could I define the number of correct and wrong numbers
> in collaboration with their amount?For example:
> The right answers are 8 and wrong are 2
> The right answers are 2 4 6 8 10 12 14 16
> The wrong answers are 22 44
> 

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Re: C program help

[vignesh1988i]

           1) SEE , wat i feel is that you can have a count after u answer
for the question u ask using printf statement... that count must be
initilized to 0 which will act as a question number. . 
           2) And since u are a beginner in C , iam suggesting u to make use
of two arrays where u can have one for right answers and other for wrong
answers.... 
           3) In the array of right answers  where u are checking ur answer
== x+x , inside that statement assign the count to the right answerd array
as:
                              right[i]=count;
           4) if ur if becomes false , in else same u must do as above but
that should be :
                              wrong[j]=count;
after this u would get ur desired output....... any doubts , mail me
:softvig88@gmail.com
thank u
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