not looping after if()

I received a chain email (arghhh) recently and I thought I would try and make a program out of this particular one for practice.

The point of the program is to figure out the pattern for a few math riddles and and use the pattern to answer the final math question.

Code:

#include <stdio.h>

int main()
{
          int user_input, loop = 1; 
          char rept;
          {  
          printf("\nTry the following!\n");
          printf("\nIf:\n");
          printf("3 + 5 = 24\n");
          printf("5 + 4 = 45\n");
          printf("6 + 8 = 84\n");
          
          printf("\nThen:\n");
          }
          do  
          { 
            {
            printf("\n9 + 7 = ");
            loop++;
            scanf("%d", &user_input);
            
            loop = user_input;
            
            printf("\nWrong! Try Again? (y)es or (n)o: ");
            rept = getchar();
            getchar();
            }              
            if((rept == 'y') || (rept == 'Y'))
            {
                     continue;
            }
            else
            {
                     break;
            }
          }
          while(loop != 144);
          {
                     printf("\nCorrect!\n");
          }
          return 0;
}

The problem is I'm not sure what statement to use if the user selected y or Y.

Code:

            if((rept == 'y') || (rept == 'Y'))
            {
                     continue;
            }
            else
            {
                     break;
            }
          }

continue; just ends the loop and prints Correct!

I have been advised against using continue; and break; by my lecturer (what are my options?).

I was also told that break; is like force quitting the program, is that true?

The break; statement breaks out of the loop dropping through to the next statement after the

Code:

while () {}

loop ends

This block of code is what you want:

Code:

if (rept != 'y' && rept != 'Y')
  break;

But if your professor doesn't want you to use break; or continue then you have to put the condition in the while loop test condition:

Code:

unsigned int yn = 'y';
while ((yn == 'Y' || yn == 'y') && answer != number) {
....
    printf("\nWrong! Try Again? (y)es or (n)o: ");
    yn = getchar();
}

Ok, but the problem still remains that if the user wants to try again, it exits the loop.

The code does, however, need some more fixing, because at the moment even the right answer is wrong when running the program.

Try this code, it will solve your requirement.

Code:

#include <stdio.h>

int main()
{
        int user_input, loop = 1;
        char rept;
        {
                printf("\nTry the following!\n");
                printf("\nIf:\n");
                printf("3 + 5 = 24\n");
                printf("5 + 4 = 45\n");
                printf("6 + 8 = 84\n");

                printf("\nThen:\n");
        }
        do
        {
                printf("\n9 + 7 = ");
                scanf("%d", &user_input);

                loop = user_input;

                if(loop == 144 )
                        break;

                printf("\nWrong! Try Again? (y)es or (n)o: ");

                /* Here you made mistake */
                getchar(); // here newline will be omitted
                rept = getchar(); 
                

        } while((rept == 'y') || (rept == 'Y'));

        if(loop == 144)
                printf("\nCorrect!\n");
        return 0;
}

And some of the changes I made in your code.
Or you can check and print inside the loop itself.

Code:

if(loop == 144)
{
printf("Correct\n");
break;
}

If you used this, There is no need to check and print outside the loop.

I was also told that break; is like force quitting the program, is that true?

Nope, Break is used to come out from the loop , it is used to exit from the loop not from the code.