when I compile it, I see same equation keep repeating.

when I compile this code, I see same equation keep repeating. I need to get different equations that the user has requested and at the end to loop it and get what is correct out of total requested problems i.e 3/5. I can't figure out to get the percentage of what they got too. Please help.thanks.

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<time.h>

int main(void){
int count=1 , times=1, x, number;
int num1, num2, op;
float answer, answer2 = 0;

srand(time(0)); op = rand()%4;
num1 = rand()%101-50;
num2 = rand()%101-50;

printf("Welcome, how many problems would you like?\n");
scanf("%d",&x);
for (times; times <= 1; times += 1) {
for (count; count <= x; count += 1) {
printf("Here is the random problem number %d \n", count);

if(op == 0) /*this is where the operations start*/
{
answer = num1 + num2;
printf("%d + %d=?\n\n", num1, num2);
printf("\nAnswer:");
scanf("%f", &answer2);
}
else if (op == 1)
{
answer = num1 - num2;
printf("%d - %d=?\n\n", num1, num2);
printf("\nAnswer:");
scanf("%f", &answer2);
}
else if (op == 2)
{
answer = num1 * num2;
printf("%d * %d=?\n\n", num1, num2);
printf("\nAnswer:");
scanf("%f", &answer2);
}
else if (op == 3)
{
answer = num1 / num2;
printf("%d / %d=?\n\n", num1, num2);
printf("\nAnswer:");
scanf("%f", &answer2);
}
}
while (answer != count) {
printf("\nYou got %d correct. That is %d!\n");
scanf("%f", &answer);
}
}
getchar();
}

---

Yech, sorry, but this formatting is terrible. Here's an example of what it should look like:

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<time.h>

int main(void)
{
    int count=1 , times=1, x, number;
    int num1, num2, op;
    float answer, answer2 = 0;

    srand(time(0));
    op = rand()%4;
    num1 = rand()%101-50;
    num2 = rand()%101-50;

    printf("Welcome, how many problems would you like?\n");
    scanf("%d",&x);
    for (times; times <= 1; times += 1)
    {
        for (count; count <= x; count += 1)
        {
            printf("Here is the random problem number %d \n", count);

            if(op == 0) /*this is where the operations start*/
            {
                answer = num1 + num2;
                printf("%d + %d=?\n\n", num1, num2);
                printf("\nAnswer:");
                scanf("%f", &answer2);
            }
            else if (op == 1)
            {
                answer = num1 - num2;
                printf("%d - %d=?\n\n", num1, num2);
                printf("\nAnswer:");
                scanf("%f", &answer2);
            }
            else if (op == 2)
            {
                answer = num1 * num2;
                printf("%d * %d=?\n\n", num1, num2);
                printf("\nAnswer:");
                scanf("%f", &answer2);
            }
            else if (op == 3)
            {
                answer = num1 / num2;
                printf("%d / %d=?\n\n", num1, num2);
                printf("\nAnswer:");
                scanf("%f", &answer2);
            }
        }
        while (answer != count)
        {
            printf("\nYou got %d correct. That is %d!\n");
            scanf("%f", &answer);
        }
    }
    getchar();
}

---

And don't forget to "return 0;" at the end!

Anyway, regarding the problem you described: "op" is given a random value at the beginning of the program, before you enter any of your loops. This value is never updated after that, so it will always pick the same equation for each iteration of the loop. Perhaps if you updated the "op" variable within the second "for()" loop, you'd get a different result each time.

Code:

---

printf("\nYou got %d correct. That is %d!\n");

---

And you aren't providing variables to match the two format specifiers in this "printf()" statement.

So much for paying attention.
http://cboard.cprogramming.com/c-pro...ml#post1112314

As Matticus said you are not updating the value of 'op' so it will stay te same every time the for loop starts

Here is my code, division needs to be 2 decimals, also I can't figure out how to input the right code for total correct and wrong answers and the right code for what percentage was attained. thanks.

//included libraries

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<time.h>

int main(void)
{
    int count=1 , times=1, x, number;
    int num1, num2, op;
    float answer, answer2 = 0;

    srand(time(0));



    printf("Welcome, how many problems would you like?\n");
    scanf("%d",&x);
    for (times; times <= 1; times ++)
    {
        for (count; count <= x; count ++)
          {
            printf("Here is the random problem number %d \n", count);
            op = rand()%4;
            num1 = (rand()%101)-50;
            num2 = (rand()%101)-50;
          {

              if(op == 0) /*this is where the operations start*/
              {
                answer = num1 + num2;
                printf("%d + %d=?\n\n", num1, num2);
                printf("\nAnswer:");
                scanf("%f", &answer2);
              }

              else if (op == 1)
              {
                answer = num1 - num2;
                printf("%d - %d=?\n\n", num1, num2);
                printf("\nAnswer:");
                scanf("%f", &answer2);
              }
              else if (op == 2)
              {
                answer = num1 * num2;
                printf("%d * %d=?\n\n", num1, num2);
                printf("\nAnswer:");
                scanf("%f", &answer2);
              }
              else if (op == 3)
              {
                answer = num1 / num2;
                printf("%d / %d=?\n\n", num1, num2);
                printf("\nAnswer:");
                scanf("%f", &answer2);
              }

        while (answer!=answer2)
            {

            printf("\nYou got %d/%d correct. That is %d%!\n", answer, answer2);

            }
          }

        }

    }

    return 0;
}

---

Well at least it has some indentation now.

> printf("\nYou got %d/%d correct. That is %d%!\n", answer, answer2);
Look at this line
1. Count the % conversions, and look how many parameters you're passing.
2. Look at each type of conversion (say %d), and cross-reference with the types of the variables (say double). You can't print a double using %d.

Thanks Salem, still learning to be like you guys.

Total correct answers over total problems, then figure the percentage. see attached image. also the code;

Code:

---

#include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
 #include<time.h>

int main(void)  {

  int count = 1,i, x, nCount = 1, times = 1;
  int num1, num2, op;
  srand(time(0));
  float answers = 0.0f, answers2 = 0.0f;

  printf("Welcome, how many problems would you like?\n");
  scanf("%d", &x);

  for (count; count <= x; count++)
  {
      for (count; count <= x; count ++)
      {
        printf("Here is the random problem number %d \n", count);
        op = rand() % 4;
        num1 = (rand() % 101) - 50;
        num2 = (rand() % 101) - 50;
        if (op == 0) {
      /*this is where the operations start */
   
      answers = num1 + num2;
      printf("%d + %d=?\n\n", num1, num2);
      printf("\nanswer:");
      scanf("%f", &answers2);
      }
    else if (op == 1)
    {
      answers = num1 + num2;
      printf("%d - %d=?\n\n", num1, num2);
      printf("\nanswer:");
      scanf("%f", &answers2);
      }
    else if (op == 2)
      {
      answers = num1 * num2;
      printf("%d * %d=?\n\n", num1, num2);
      printf("\nanswer:");
      scanf("%f", &answers2);
      }
    else if (op == 3)
      {
      answers = roundf((((num1 * 1.0f) / num2)) * 100.0f)/100.0f;
      printf("%d / %d=?\n\n", num1, num2);
      printf("\nanswer:");
      scanf("%f", &answers2);
      answers2 = roundf((answers2) * 100.0f)/100.0f;
      }
  }
  for (count = 0; count < times; count++)
      for (times=0; times<=x; times++)
      times = times + 1;


    printf("\nYou got %0.0f out of %0.0f correct. That is %d percent!\n",  count++, x);

  return 0;
}

---

Attachment 11760

>> printf("\nYou got %0.0f out of %0.0f correct. That is %d percent!\n", count++, x);

Three conversions, two numbers. You've already been told to pass the correct number of arguments. Also make sure that the types of the variables match the conversions.

I think you're doing the percentage completely wrong: part / whole * 100

Your code compiles!?

Code:

---

$ gcc -Wall -Wextra -ansi foo.c
foo.c: In function `main':
foo.c:48: warning: implicit declaration of function `roundf'
foo.c:60: warning: double format, different type arg (arg 2)
foo.c:60: warning: double format, different type arg (arg 3)
foo.c:60: warning: too few arguments for format
foo.c:60: warning: double format, different type arg (arg 2)
foo.c:60: warning: double format, different type arg (arg 3)
foo.c:60: warning: too few arguments for format
foo.c:63: error: parse error at end of input
foo.c:8: warning: unused variable `i'
foo.c:8: warning: unused variable `nCount'

---

It's missing a closing brace.

See all those warnings on line 60.
That's due to all the inconsistencies in your printf call.

You would do yourself a favour if you got the gcc compiler (or an IDE that comes with gcc pre-configured, such as code::blocks). Then you'll be able to check for a lot more suspect things in your code before running it (only to find it doesn't work).

What is the purpose of this?

Code:

---

  for (count; count <= x; count++)
  {
      for (count; count <= x; count ++)

---

If you just deleted one of them, I suspect things would be a lot better.
You should say
for (count=1; count <= x; count ++)

Also mystified by this:

Quote:

---

Originally Posted by KumarSingh

Code:

---

  for (count = 0; count < times; count++)
      for (times=0; times<=x; times++)
      times = times + 1;

---

---

No idea what this is meant to be doing. Pretty sure it's not doing whatever it's meant to.

To work out the percentage you need to know the total number of questions (which you do: it's x) and the number of correct answers (which you don't). You need to check the answers given are right and keep count of the number correct. I'd focus on that and worry about the percentage later.

Code:

---

answers = roundf((((num1 * 1.0f) / num2)) * 100.0f)/100.0f;

---

You can use a cast rather than multiplying by 1.0f to get a float type:

Code:

---

answers = roundf(((float)num1 / num2) * 100.0f)/100.0f;

---

That's quite a nifty way of getting 2 decimal places. I like!
Beware that roundf() isn't standard though (I think it's gcc specific) so if this needs to be compilable with other compilers you should fine an alternative way of rounding.

I removed that part

Code:

---

for (count; count <= x; count ++)

---

and it doesn't change anything. the only part am stuck is getting the right code for right answers/total answers and the then the code for what percentage was correct.