Thread: Program stops at first input

For starters, you should initialize your num array before ever referencing it. Your while loop is referencing a slot in your array and you don't even know what value will be there.

Secondly, your first for loop is an infinite loop. numDigits is not being changed inside the body of the loop anywhere, so as long as numDigits!=0 the loop will continue.

Hello,

ok here are some suggestions so you can can become better at programming.

- I do not like to declare variables in a single line because it's hard to read but that it just my prefference. When I change it to a per line basis this shows up:

Code:

int num[1000];
int sum = 0;
int i = 0;
int j = 0;
int numDigits;
int counterDividible = 0;
int counterDigits = 0;

Such a big code of variables might tell you that the procedure is trying to do a lot. This bing us to the other section.

- Write pseudocode it will guide you so you do not end up with a huge single procedure and you devide it into meaningfull parts. Lets get the pseudocode for your task(s)

1. Enter numbers until 0 or -1
2. Get sum of all positive and 2 digit numbers
3. Get the count of divisible by 3.

This pseudocode helped us correctl identify several functions:

Code:

int isTwoDigitNumber(int number);
int isDivisibleByThree(int number);

- Use constants and enumerators to make your whole code read easily

Code:

enum {NO_ERRORS, ERROR};
enum {FALSE, TRUE};

Finally after all this we have something like the code below. I have left the function implementations for you to do.

Code:

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

/* constants and enumerators */
enum {NO_ERRORS, ERROR};
enum {FALSE, TRUE};

/* function prototypes */
int isTwoDigitNumber(int );
int isPossitive(int );
int isDivisibleByThree(int );

int main(void) {
    /* Enter numbers until you enter 0 or -1
    and print sum of all positive two-digit numbers entered
    and print how many of these numbers divisible by 3 */

    int sum = 0;
    int number = 0;
    int counterDivisible = 0;
    int leave = FALSE;

    while (!leave){
        scanf ("%d", &number);

        if (number == 0 || number == -1){
            leave = TRUE;
        }else{

            if (isPossitive(number) && isTwoDigitNumber(number)){
                sum += number;
            }

            if (isDivisibleByThree(number)){
                counterDivisible++;
            }
        }
    }

    printf("Sum: %d \n Divisible by 3:%d", sum , counterDivisible);
    return NO_ERRORS;

}


/* checks if number has 2 digits then TRUE else FALSE */
int isTwoDigitNumber(int number){

}


/* checks if number is possitive then TRUE else FALSE */
int isPossitive(int number){

}


/* checks if number is divisisible by 3 then TRUE else FALSE */
int isDivisibleByThree(int number){

}

Hope this helps

Now you're incrementing j just before the condition is checked, so that num[j] is the element after the one you scanf'd the number into.

Your while condition seems to be missing a !.

Code:

do{         
        scanf("%d", &num[j]);
         
        if(num[j] < 0){
            num[j] = -num[j];
        }
        numDigits = num[j];

         
        for( ; counterDigits != 0; counterDigits++){
            counterDigits = numDigits / 10;
        }
         
        if( counterDigits == 2){
             
            sum += num[j];
        }
         
        j++;//←remove this line

         
         
    } while( (!num[j] == 0) || !(num[j] == -1) ); // this is probably what you meant to say here.

Use prints to troubleshoot your code yourself. When that do-while terminates after the first iteration, you should know that its condition is somehow evaluating to false when you don't expect it to. Printing out both parts of that condition would tell you which one is not doing what you expected. That would expose one of the problems. Doing the same thing above line 31(j++;) would expose the other problem if you hadn't noticed it by then.

Code:

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


int main(void) {


   
	int num, sum = 0 , i = 0, counter3 = 0, check;
	do{
		
		scanf("%d", &num);
		
		check  = num;
		for( i = 0; check  != 0; i++){
			
			check  = check / 10;
			
		}
		
		if( i == 2){
			sum = sum + num;
		}
		if(abs(num) % 3 == 0){
			counter3++;
		}
		
	}while( num != 0 && num != -1 );
	
	printf("Sum: %d\nDividble by 3: %d", sum , counter3);


    return 0;
   }

Did it without arrays it should be okay now ? Trying to find a bug.

I recommend that you redo from scratch. This time, write a bit, compile and test, then fix the bugs. When you are satisfied the small part you wrote works, write a bit more, compile and test, etc.

In particular, handle this first:
Enter numbers until you enter 0 or -1
Notice that you do not need an array for this. Of course, just entering numbers would not be very informative, so you could arrange to print each number after it is entered, except for the terminating value of 0 or -1.

Once you have a loop that works perfectly, do this:
and print sum of all positive two-digit numbers entered
Notice again that you still do not need an array: you do need to identify the positive two-digit numbers (fredlo2010's suggestion of a function is fine, though it is sufficiently trivial that you don't need one) one by one, but you only need a variable to keep track of their sum.

Finally, after you have tested and are sure that everything works, do this:
and print how many of these numbers divisible by 3
Again, notice that an array is not needed: you just need to identify those numbers that are divisible by 3, one by one, and then have a counter to keep track of how many of them there are.

Originally Posted by laserlight

I recommend that you redo from scratch. This time, write a bit, compile and test, then fix the bugs. When you are satisfied the small part you wrote works, write a bit more, compile and test, etc.

In particular, handle this first:
Enter numbers until you enter 0 or -1
Notice that you do not need an array for this. Of course, just entering numbers would not be very informative, so you could arrange to print each number after it is entered, except for the terminating value of 0 or -1.

Once you have a loop that works perfectly, do this:
and print sum of all positive two-digit numbers entered
Notice again that you still do not need an array: you do need to identify the positive two-digit numbers (fredlo2010's suggestion of a function is fine, though it is sufficiently trivial that you don't need one) one by one, but you only need a variable to keep track of their sum.

Finally, after you have tested and are sure that everything works, do this:
and print how many of these numbers divisible by 3
Again, notice that an array is not needed: you just need to identify those numbers that are divisible by 3, one by one, and then have a counter to keep track of how many of them there are.

Big thank you @laserlight you always have constructive and informative answers that really help.

Code:

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


int main(void) {


  int num, sum = 0 , i = 0, counter3 = 0, check;    

  do{
        
        scanf("%d", &num);
        check  = num;
        for( i = 0; check  != 0; i++){
            
            check  = check / 10;
            
        }
        
        if( i == 2){
            sum = sum + num;
        }
        if(abs(num) % 3 == 0 && num != 0 && num != -1){
            ++counter3;
        }
    }while( num != 0 && num != -1);
    
    printf("Your sum:%d\nDividible by 3: %d", sum, counter3);


    return 0;
   }

Works like a charm