Thread: Checking if an integer is a Prime number program too many prints

Code:

#include <stdio.h>

Code:

#include <math.h>


int main(void)
{
	int n;
	int x;
printf("enter a number to check if it is prime\n");
	scanf("%d",&x);	
    
	
	for (n=2;n<=sqrt(x);n++) {
		if((x%n)==0)
		{
printf("%d not a prime number\n",x);
		}
	    else
		{
printf("%d prime number\n",x);
		}
     } 
}

this program supposed to check if a given number prime or not,
when I enter a value bigger than 9 it gives multiple prints, moreover if all those prints say "prime number" yes it is a prime number in real, but if at least one of says "not a prime number" it is not a prime number in real
I tried many things it didn't work. I tried to find a similar code but I couldn't

try having a flag that starts out as 1 (true).
int prime = 1;
remove the printfs from your loop. instead if you find an instance of x%n == 0, set the flag to 0 (false) and break out of the loop (since one fail is enough to say not prime)
then after the loop test your flag and print that it is either prime or not prime.

after testing this, casting int as a double doesnt make a difference to it at all....surprising i know!!!

BUT doing the

Code:

int prime=1;

and using an IF to check if prime=0 to display "is prime" or "is not prime" works perfect!

your code doesnt show anything for 1,2,or 3

"new" code says
1 is prime, 2 is prime 3 is prime....etc...

Originally Posted by dmh2000

try having a flag that starts out as 1 (true).
int prime = 1;
remove the printfs from your loop. instead if you find an instance of x%n == 0, set the flag to 0 (false) and break out of the loop (since one fail is enough to say not prime)
then after the loop test your flag and print that it is either prime or not prime.

Code:

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


int main(void) {
    int n;
    int i;
    int flag;
    
printf("Enter value of N > ");
    scanf("%d", &n);
    flag = 1;
    for (i=2; (i<sqrt(n));i++) { 
        
        if ((n % i) == 0) 
            flag = 0;
        
            
    }
    
    if (flag=0)
printf("%d is prime\n", n);
else
printf("%d is not prime\n", n);
return0;
}

I came up with this but still not working

I haven't analyzed your code, but a quick look-over brought this to my attention:

Code:

if (flag=0)

You mean ==, not =

curious why you changed x to i?

also you changed your code from

Code:

for (n=2;n<=sqrt(x);n++) {
        if((x%n)==0)

to

Code:

    for (i=2; (i<sqrt(n));i++) { 
        
        if ((n % i) == 0)

which changes your math, and what is devided by what

other then that, your code matches mine (to a point!!!)

//edit//
ps although on some math formulas you need to cast ints to doubles (floats) for this you dont, and it will work

ok now for the probs!!!

Code:

for (i=2; (i<sqrt(n));i++) {

should have been

Code:

for (i=2; (i<=sqrt(n));i++) {

when you "fixed" it, you missed the =

your

Code:

return0;

should be

Code:

return 0;

but your biggest mistake, is your flag

Code:

    if (flag=0)

you not only missed the second =

but your flag would be backwards!!!

flag of 1 says it IS a prime
flag of 0 says it is not prime!!!

Code:

    if (flag==1)

not too bad for not doing it as long as most of us!

Code:

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


int main(void) {
    int n;
    int i;
    int flag;
    
printf("Enter an integer to check if it is prime: ");
    scanf("%d", &n);
    flag = 1;
    for (i=2; (i<=(int)sqrt(n));i++) { 
        
        if ((n % i) == 0) 
            flag = 0;
    }
    
    if (flag==1)
printf("%d is prime\n", n);
else
printf("%d is not prime\n", n);
return0;
}

here it is finally working fine but I am still not sure how this flag works. I mean how copiler process this flag?
Thanks to everybody who helped me out here!

...but I am still not sure how this flag works.

Take out a pen and paper. Pick out a value for 'n' and write it down. Then, by hand, step through each line of code, writing out all necessary values. Follow it through each step to see what your program is doing.

It's good you got it to work, but it seems that's only the case because you received answers here. When you write a program yourself, you should fully understand what each and every line of code is doing. Take this advice if you really want to better understanding programming, and become proficient in it.

Code:

#include <stdio.h>

Code:

#include <math.h>


int main(void) {
    int n;
    int i;
    int flag;
    
printf("Enter an integer to check if it is prime: ");
    scanf("%d", &n);
    flag = 1;                               /* flag value is 1(true) now */
for (i=2; (i<=(int)sqrt(n));i++) {      /* giving "i" integer values increasing one by one (i++)
                                             starting from 2 to till equals to square root of "n"*/
        
if ((n % i) == 0)                   /* divide n by for each "i" and check the remainder for each "i" and if any remainder 
                                             equals to zero then flag is 0(false  */
            flag = 0;
    }
    
    if (flag==1)
printf("%d is prime\n", n);         /*check flag for each "i" values if it is 1(true) and print n is "prime"*/
else
printf("%d is not prime\n", n);     /*all others ( if it is 0(false) print "not prime"*/
return0;
}

I placed the comments next to lines, thats how I understood compiler do the process for this code pls tell me if I thought wrong

You have to work on your code posting skills. Just copy your code as plain text to the text box and put code-tags around it.
Your indentation is completley messed up too (which could be related to the above problem).

Code:

if (flag==1) 
    printf("%d is prime\n", n);         /*check flag for each "i" values if it is 1(true) and print n is "prime"*/ 
else 
    printf("%d is not prime\n", n);     /*all others ( if it is 0(false) print "not prime"*/

You don't check the flag for each "i". You check the flag after the loop whether any value of "i" has set it to 0.

Just some additional comments:
1) You calculate the square root for "n" at each iteration although "n" doesn't change. Therefore it's better to calculate it once before the loop and then just use the stored value.
2) If your test fails when "i" is 2 (i.e. "n" is odd) all further even numbers (4, 6, 8, ...) will fail. Thus you should only loop through odd values for "i"
3) As soon as you find a value for "i" which divides "n" there is no need to try other values. You can just break out of the loop.
4) If you do break out prematurely there is really no need for a flag. Just test if "i" is equal to the square root of "n" after the loop to decide whether "n" is prime.

Bye, Andreas

Originally Posted by AndiPersti

You have to work on your code posting skills. Just copy your code as plain text to the text box and put code-tags around it.
Your indentation is completley messed up too (which could be related to the above problem).

Code:

if (flag==1) 
    printf("%d is prime\n", n);         /*check flag for each "i" values if it is 1(true) and print n is "prime"*/ 
else 
    printf("%d is not prime\n", n);     /*all others ( if it is 0(false) print "not prime"*/

You don't check the flag for each "i". You check the flag after the loop whether any value of "i" has set it to 0.

Just some additional comments:
1) You calculate the square root for "n" at each iteration although "n" doesn't change. Therefore it's better to calculate it once before the loop and then just use the stored value.
2) If your test fails when "i" is 2 (i.e. "n" is odd) all further even numbers (4, 6, 8, ...) will fail. Thus you should only loop through odd values for "i"
3) As soon as you find a value for "i" which divides "n" there is no need to try other values. You can just break out of the loop.
4) If you do break out prematurely there is really no need for a flag. Just test if "i" is equal to the square root of "n" after the loop to decide whether "n" is prime.

Bye, Andreas

Yea I know I will work on my indentations...
Thanks for suggestions I think I can do top 3 easily, but I am not sure about breaking out prematurely I will try though