Prime Number program - c

Question

I want to make a program in C language which will take the user input and I would not be able to understand the logic of the loop.
for ( c = 2 ; c <= n - 1 ; c++ )
The program code is given below:-
#include<stdio.h>
#include<conio.h>
void main()
{
int n, c;
printf("Enter a number to check if it is prime\n");
scanf("%d", &n);
for ( c = 2 ; c <= n - 1 ; c++ )
{
if ( n % c == 0 )
{
printf("%d is not prime.\n", n);
break;
}
}
if ( c == n )
printf("%d is prime.\n", n);
getch();
}
I have used the for loop which will end up the statement of n - 1 in for loop. If I will give the input 11 then it will end up on 11 - 1 = 10 then how it will give up the logic of if(c == n) { printf("%d", n);?

Answer 1

If I will give the input 11 then it will end up on 11 - 1 = 10 then how it will give up the logic of if(c == n) { printf("%d", n);?
Now correctly understand your for loop condition:
for ( c = 2 ; c <= n - 1 ; c++ )
^^^^^^^^^^^^
2 <= 11 - 1 -> True // {for block executes }
3 <= 11 - 1 -> True // {for block executes }
:
:
9 <= 11 - 1 -> True // {for block executes }
10 <= 11 - 1 -> True // {for block executes }
11 <= 11 - 1 -> False breaks //{And Now for block NOT executes}
if (c == n)
^^^^^^
11 == 11 -> True // {if block executes}
According to for loop condition c <= n - 1, loop breaks when c value becomes equals to n. So if n is equals to 11 loop condition is true for c = 2 to c = 10, in each iteration c increments by one (using c++ increment) when c becomes 11 (ot say n) then condition c <= n - 1 become false and loop breaks.
In if condition (after for loop) c value compared with n. that is:
if ( c == n )
// 11 == 11 that is true
for n = 11 it becomes and c = 11 if condition evaluates true and printf() associated with if executes.
It is also important to understand that the for-loop only terminates for c = n when n is a prime number, but if suppose n is a non-prime number then for-loop will break for c value less then n - 1 due to break; statement in nested if block in for-loop.
for( c = 2; c <= n - 1; c++ )
{
if(n % c == 0)<=="for Non-prime `n`, if condition will be true for some `c < n - 1`"
{ ^^^^^^^^^^^ True
printf("%d is not prime.\n", n);
break; <== "move control outside for-loop"
} // |
} // |
// <---------+ // if break; executes control moves here with c < n - 1
if (c == n)<== "this condition will evaluates FALSE"
^^^^^^^^ False
For example if n = 8 then in very first iteration of for-loop with value c = 2 if condition if(n % c == 0) that evaluates as if(8 % 2 == 0) == if( 0 == 0) = True and break; statement inside if-block moves control outside for-loop(as shown in figure).
Because this time for loop not terminated due to c <= n - 1 condition but braked because of if(n % c == 0) so out-side for-loop c value is less than n hence if (c == n) evaluates as False.

Answer 2

The for loop loops from c = 2 to c = n - 1 except it hit's the break statement. If it does, it will jump out of the loop. If you never break then your c will actually be n after the loop.
And here is why. The loop works like this:
initialize the for loop with c = 2
Check for condition (c <= n - 1)
if true: execute loop body
if false: jump past loop
increment c by one
goto 2.
Example: Suppose your n is 3.
we set c to 2, now c == 2 and n == 3
2 <= 3 - 1 is true, so the loop body will be executed
we increment c by 1, now c == 3 and n == 3
we go back to 2. in the description
3 <= 3 - 1 is false, so we don't execute the loop body now and jump out of the loop
after we left the loop c == 3 and n == 3 so c == n
So if we never hit the break statement c will be equal to n after the loop. We hit the break statement if n is not prime. If we break c will miss at least one increment and thus c < n after the loop. Now c == n will evaluate to false and the if statements body of if ( c == n ) will not be executed.
Now to the if ( n%c == 0 ). n%cmeans n modulo c, so the remainder of the division of n by c. If this remainder is 0 then c is an integer divisor of n. So in the loop you are testing n for any divisor bigger than 1 and smaller than n.
If there is a divisor of n except 1 and itself, n can't be prime. So if you hit any c with 1 < c < n that makes n%cequal to 0 n can't be prime.
Hint: You don't have to test for divisors bigger than √n.

Answer 3

Your assumption that c ends up being n - 1 is incorrect. If you step through your program with the debugger you should see that for n == 11, c == 11 at the end of the loop.
If we don't break early then the last time the loop body executes is indeed when c == n - 1 however c is then incremented and the loop invariant test fails so after the loop c == n.

Answer 4

Consider what it means when you say that a number is prime.
A number p is prime when for every n>1, n<p, the remainder r = 0 for p/n = r.
So, this loop runs through each value (c) the range [2..p-1], testing for the remainder.
int isprime = 1; //use a flag to indicate prime, assume prime until proven otherwise
for ( c = 2 ; //initialize c=2, the first value in the range [2..p-1]
c <= n - 1 ; //check whether c is still in the range [2..p-1]
c++ ) //increment c to the next value in the range [2..p-1]
{
if ( n % c == 0 ) //modulo '%' is the remainder of the integer division
{
isprime = 0;
break;
}
}
printf("%d %s prime.\n", n, isprime?"is":"is not");
Rather than test that the number is prime by checking a side effect of the loop counter, why not set a flag before the loop, and then test it as the decision for prime at the end? The result is code that is less fragile, error-prone, and clearer.
Want to save about half the work? Once you test n%2, we know that there is no number k such that k*2=n, right? so check the range [2..p/2], and change your loop to,
for( c = 2; c <= n/2; c++ )
Can you think of a number smaller than n/2 that works?
An interesting algorithm for finding multiple primes is the Sieve of Erastosthenes.