To convert a 'for' loop into a 'do..while' loop

Question

I came across a program to print all the prime numbers between 1 to n, where 'n' is the value supplied by the user. It used the for loop. The program is as follows-

#include <stdio.h>

int main()
{
    int i, j, end, isPrime; // isPrime is used as flag variable

    /* Input upper limit to print prime */
    printf("Find prime numbers between 1 to : ");
    scanf("%d", &end);

    printf("All prime numbers between 1 to %d are:\n", end);

    /* Find all Prime numbers between 1 to end */
    for(i=2; i<=end; i++)
    {
        isPrime = 1; 
        for(j=2; j<=i/2; j++)
        {
            if(i%j==0)
                isPrime = 0;
        }
        if(isPrime==1)
        {
            printf("%4d", i);
        }
    }
    return 0;
}

I very well understand the above code. But just to test my knowledge , I tried writing the same program using the do loop. It didn't work out very well. I searched through books and internet to find a program where they use a do loop to compute prime numbers till n. But couldn't find one. I wanted to know if its even possible or not. To show my efforts, I put below the code I made-

#include <stdio.h>
main()
{
    int i, j, n, isPrime;
    printf("Enter n\n");
    scanf ("%d", &n);
    i = 2;

    do 
    {
        do
        {
            j = 2;
            if (i%j == 0)
                isPrime = 0;
            else
            {
                isPrime = 1;
                printf ("%d", i);
            }
            j = j+1;
        }
        while (j <= i/2);

        i = i+1;
    }
    while (i <= n);
}

Answer 1

You had a good start, but made a few small errors which resulted in big problems

j=2 was set in the second do-while loop. This resulted in j being set to 2 every loop, thus never exiting the do while loop.

Next was the if/else. In the for loop example, the if (isPrime==1) was outside the for loop. You were checking inside the while loop resulting in things being named prime way to soon.

Below is an example of the working code

#include <stdio.h>
main()
{
    int i, j, n, isPrime;
    printf("Enter n\n");
    scanf ("%d", &n);
    printf("All prime numbers between 2 to %d are:\n", n);

    i = 2;
    do 
    {
        isPrime = 1;
        j = 2;

        do
        {
            if (i%j == 0)
                isPrime = 0;
            j = j+1;
        } while (j < i/2);

        if (isPrime == 1 || i == 2)
            printf ("%d\n", i);
        i = i+1;
    }
    while (i <= n);
}

Answer 2

Try this code instead:

#include <stdio.h>
main(){
int i, j, n, isPrime;
printf("Enter n\n");
scanf ("%d", &n);
i = 2;
do 
{
    isPrime = 1;

    j = 2;

    do
    { 
        if (i%j == 0 && j <= i/2)
            isPrime = 0;
        j = j+1;
    }
    while (j <= i/2);

    if(isPrime == 1) printf ("%d\n", i);

    i = i+1;
}
while (i <= n);
}

There is difficulty in translating a for loop to a do-while loop.

For example, with problems you had in your code:

• At the start of the second do loop, you were setting j to 2, over and over. This meant that you were stuck in an infinite loop, since at the start of the loop, the value of j would be reset.
• isPrime should be set as 1 at the start of the first do-while loop. This is so that, if i is, at any point divisible by j, the flag is set to 0, meaning i is not prime. The problem with the way you set it is that if the last value of j does NOT divide into i, then isPrime is set to 1, even if previous values of j divided into it.

Answer 3

A for loop

for (INIT; CONDITION; NEXT) {
    BODY;
}

is equivalent to the while loop

{
    INIT;
    while (CONDITION) {
        BODY;
        NEXT;
    }
}

(except that the for loop does not allow declaring the same name in the INIT and in the outermost block of the BODY).

You made a couple of mistakes in your conversion:

1. The do { BODY; } while (CONDITION); syntax always executes the BODY at least once. The original for loop might not.

This matters for the case i==2: In the original code, the inner for loop will set j=2, then immediately test whether j<=i/2, which is 2<=1 or false, so the loop body won't execute at all. But your code starts by setting i=2; j=2; then going right to the factor test if (i%j == 0). This is true, so it incorrectly concludes 2 is not a prime.

2. You moved the isPrime = 1; and printf statements from the outer loop over i to be inside the inner loop over j. This changes the program's behavior in a bad way.

Consider what happens when i==15: We start with j=2, and find that i%j != 0 since 2 is not a factor of 15. So it sets isPrime = 1; and prints 15. The inner loop continues, setting j=3. This time i%j == 0 since 3 is a factor of 15, so it sets isPrime = 0; and prints nothing. The inner loop goes again with j=4. Now it's back to i%j != 0, so it sets isPrime = 1; and prints 15 again. This continues up to j=7, printing 15 a total of four times (when j is 2, 4, 6, and 7).

I'll leave it to you to see if you can now correct the code using while loops.

Answer 4

here is the code that you need.

#include <stdio.h>
main() {
  int i, j, n, isPrime;
  printf("Enter n\n");
  scanf("%d", &n);
  i = 2;
  do {
    j = 2;
    isPrime = 1;
    if (i == 2)
      printf("%d ", i);
    do {
      if (i % j == 0) {
        isPrime = 0;
        break;
      }
      j = j + 1;
    } while (j <= i / 2);
    if (isPrime)
      printf("%d ", i);
    i = i + 1;
  } while (i <= n);
}