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

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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);
}
```

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);
}
```

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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);
}
```

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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`).

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.

1. 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.

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