Code to count Prime numbers between a range returns half of the numbers

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Some piece of this code is wrong and I can't figure it out. I write the code logically to count Prime Numbers between two given numbers but the code returns half of the range of given numbers. The functions works fine but the logic of while loop is wrong

The code:

#include<stdio.h>

int checkPrime(int isItPrime);

int main()
{

    printf("Counting Prime Numbers between twos\nEnter two numbers , the First number must be smaller\n");
    int num1, num2;
    scanf("%d%d", &num1, &num2);
    int count = 0;
    while(num1<num2)
    {
        num1++;
        if(checkPrime(num1)==1)
            count++;
    }
    printf("%d\n", count);

    return 0;
}

int checkPrime(int isItPrime)
{

    int result = 0, j = 2;
    while(result < isItPrime)
    {
        result = (isItPrime%j);
        j++;
        if (result==0)
            return 0;
        else if (result!=0)
            return 1;
    }

}

See if this works,I hope it does:

 #include<stdio.h>

int checkPrime(int n);

int main()
{

    printf("Counting Prime Numbers between twos\nEnter two numbers , the First number must be smaller\n");
    int num1, num2;
    scanf("%d%d", &num1, &num2);
    int count = 0;
    while(num1<num2)
    {
        num1++;
        if(checkPrime(num1)==1)
            count++;
    }
    printf("%d\n", count);

    return 0;
}
int checkPrime(int n)
{

    if (n <= 1)
        return 0;
    for (int i = 2; i < n; i++)
        if (n % i == 0)
            return 0;

    return 1;
}

Count Primes, Count the number of prime numbers less than a non-negative number, n. divisible by any number > n / 2, we can immediately cut the total iterations half by dividing return count; } private boolean isPrime(int num) { if (num <= 1) return false; 2. public: 3. int countPrimes(int n) {. 4. 5. } 6. }; Console. Contribute. Run Code Given an array arr[] of Prime Numbers and a number M, the task is to count the number of elements in the range [1, M] that are divisible by any of the given prime numbers. Examples: Input: arr[] = {2, 3, 5, 7} M = 100


It's OK to return 0 the very first time you get result == 0, but don't return 1 the very first time you get result != 0. That will cause every odd number to be classified as prime. You have to check the result for every value of j up to the square-root of isItPrime. Also: as Brendan pointed out, your while conditional should be testing j, not result. You could say while(j*j <= isItPrime)

Also also: note that your current main() logic tests numbers excluding num1 but including num2—verify that that is in fact what you intended.

More generally: be skeptical of your code. Don't just look at the number of (alleged) primes it reports. In initial (debugging) versions of the program, you should be actually printing out each number as it's tested, along with the test result (i.e. whether it thinks each number is prime). If you had done that, you would probably have caught the problems above, or at least narrowed it down.

Print prime numbers in a given range using C++ STL, Queries for the difference between the count of composite and prime numbers in a given range. Article Tags : C++ � Mathematical � GE � Prime� Count the number of prime numbers less than a non-negative number, n. Example: Input: 10 Output: 4 Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.


OP's checkPrime() incorrectly always return in its first iteration. It need to test more values when result != 0.

int checkPrime(int isItPrime) {
    int result = 0, j = 2;
    while(result < isItPrime) {
        result = (isItPrime%j);
        j++;
        if (result==0) return 0;
        // else if (result!=0) return 1;   // problem
    }
    return 1;  // add
}

Fixed OP's algorithm is very slow as it iterates up to isItPrime when iteration only up to sqrt(isItPrime) is needed. Also the return value incorrect with values < 2.

int checkPrime_fix1(int isItPrime) {
    int result = 0, j = 2;
    // Avoid sqrt(isItPrime). Floating point has too many subtle problem for an integer task
    // Avoid j*j <= isItPrime to prevent overflow with large isItPrime
    while(j <= isItPrime/j) {
        result = isItPrime%j;
        j++;
        if (result==0)
            return 0;
    }
    return isItPrime > 1;
}

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Prime Number From 1 to 100 Program in Java, How to display prime numbers between 1 to 100 using Java Code If the input is 17, half is 8.5, and the loop will iterate through values 2 to 8; If numberToCheck is entirely divisible by another number, we return false, and� In this example, you will learn about C program to display prime numbers between two numbers interval or in a given range by the user with and without using the function. What is prime number ? A prime number is a natural number that is divisible by 1 and itself only. For example: 2, 3, 5, 7 …


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