Prime Number Program In Tcl

1. Prime Number Program In C
2. Prime Number Programm In C++
3. Prime Number Program In Tcl
4. Tcl Contact Number
5. Tcl Technical Support Phone Number

C Program to Check Whether a Number is Prime or Not Example to check whether an integer (entered by the user) is a prime number or not using for loop and if.else statement. To understand this example, you should have the knowledge of following C programming topics. Prime numbers between 1 to 100 in C Programming Language. Ask Question -4. I want to print prime numbers between 1 to 100, I write my code like the following but when I run it, it starts printing 3,7,11,17.91 Why not the code print 2? Prime Number Check 1-100 (C)-2. Unexpectedly crash in program for finding prime numbers.

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.

The program code is given below:-

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);?

Grijesh Chauhan

Prime Number Program In C

Tayyab Gulsher VohraTayyab Gulsher Vohra

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4 Answers

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:

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:

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 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.

Grijesh ChauhanGrijesh Chauhan

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 nafter the loop.

And here is why. The loop works like this:

1. initialize the for loop with c = 2
2. Check for condition (c <= n - 1)if true: execute loop bodyif false: jump past loop
3. increment c by one
4. goto 2.

Prime Number Programm In C++

Example: Suppose your n is 3.

1. we set c to 2, now c 2 and n 3
2. 2 <= 3 - 1 is true, so the loop body will be executed
3. we increment c by 1, now c 3 and n 3
4. we go back to 2. in the description
5. 3 <= 3 - 1 is false, so we don't execute the loop body now and jump out of the loop
6. 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.

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.

Grijesh Chauhan

Prime Number Program In Tcl

jk.jk.

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.

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,

Can you think of a number smaller than n/2 that works?

An interesting algorithm for finding multiple primes is the Sieve of Erastosthenes.

ChuckCottrillChuckCottrill

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What is a Prime Number?

A prime number is a number that is only divisible by 1 or itself. For example, 11 is only divisible by 1 or itself. Other Prime numbers 2, 3, 5, 7, 11, 13, 17....

Note: 0 and 1 are not prime numbers. 2 is the only even prime number.

Java Program to check whether number is prime or not

Program Logic:

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• We need to divide an input number, say 17 from values 2 to 17 and check the remainder. If remainder is 0 number is not prime.
• No number is divisible by more than half of itself. So we need to loop through just numberToCheck/2 . If input is 17, half is 8.5 and the loop will iterate through values 2 to 8
• If a numberToCheck is completely divisible by other number, flag isPrime is set to true and the loop is exited.

Output:

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Check our program to Find Prime Numbers from 1 to 100