How many prime numbers C program - c

Question

I wrote this program to find prime numbers between 1 and 50000, and I still need to find how many prime numbers there is (I tried a lot of tricks but I did not succeed)
#include <stdio.h>
//int getValueFromUser();
void PrintListOfPrime(int value);
int main() {
int value = 23;
PrintListOfPrime(value);
return 0;
}
void PrintListOfPrime(int value) {
int ValueIsPrime; //ValueIsPrime is used as flag variable
printf("The list of primes: ");
for (int i = 2; i <= value; i++) {
ValueIsPrime = 1;
/* Check if the current number i is prime or not */
for (int j = 2; j <= i / 2; j++) {
/*
* If the number is divisible by any number
* other than 1 and self then it is not prime
*/
if (i % j == 0) {
ValueIsPrime = 0;
break;
}
}
/* If the number is prime then print */
if (ValueIsPrime == 1)
printf("%d, ", i);
}
printf("\n");
}

Answer 1

I tried a lot of tricks but I did not succeed
If OP's code takes too long to ran, iterate to the square root of i, not up to i/2.
j <= i / 2 is very slow. Use j <= i / j instead.
Form a count and increment with every prime. #gspr
if (ValueIsPrime == 1) {
printf("%d, ", i);
prime_count++;
}
Bigger change yet even faster to "find prime numbers between 1 and 50000", research Sieve of Eratosthenes

Answer 2

Hello fast answer is to create a variable in main, int totaleOfPrimes = 0; for example.
then send it by reference to the fucntion :
Function declaration : void PrintListOfPrime(int value,int* counter);
Function call : void PrintListOfPrime(value,&totaleOfPrimes);
then Increment counter befor printing :
if (ValueIsPrime == 1){
(*counter)++;
printf("%d, ", i);
}

Answer 3

There is no need to iterate the loops for all numbers between 2 and value. You should consider only 2 and odd numbers.
The function can look the following way as it is shown in the demonstrative program below.
#include <stdio.h>
static inline size_t PrintListOfPrime( unsigned int n )
{
size_t count = 0;
printf( "The list of primes:\n" );
for ( unsigned int i = 2; i <= n; i = i != 2 ? i + 2 : i + 1 )
{
int isPrime = 1;
/* Check if the current number i is prime or not */
for ( unsigned int j = 3; isPrime && j <= i / j; j += 2 )
{
/*
* If the number is divisible by any number
* other than 1 and self then it is not prime
*/
isPrime = i % j != 0;
}
/* If the number is prime then print */
if ( isPrime )
{
if ( ++count % 14 == 0 ) putchar( '\n' );
printf( "%u ", i );
}
}
return count;
}
int main(void)
{
unsigned int n = 50000;
size_t count = PrintListOfPrime( n );
printf( "\n\nThere are %zu prime numbers up to %u\n", count, n );
return 0;
}

Answer 4

Run this code in C. It will return the value of a pi(x) function. It is basically the Prime counting function:
#include <stdio.h>
#define LEAST_PRIME 2
#include <math.h>
int main() //works for first 10000 primes.
{
int lower_limit = 2, no_of_sets;
// printf("NUMBER OF SETS: ");
// scanf("%d", &no_of_sets);
int remainder, divisor = 2, remainder_dump, upper_limit; //upper limit to be specified
//by user.
int i = 1;
// printf("SPECIFY LOWER LIMIT: ");
// scanf("%d", &lower_limit);
int number_to_be_checked = lower_limit;
printf("SPECIFY UPPER LIMIT: ");
scanf("%d", &upper_limit);
printf("2\t\t\t\t", number_to_be_checked);
//PRINTS 2.*/
do
{
remainder_dump = 1;
divisor = 2;
do
{
remainder = number_to_be_checked % divisor;
if (remainder == 0)
{
remainder_dump = remainder_dump * remainder; // dumping 0 for rejection.
break;
}
++divisor;
} while (divisor <= number_to_be_checked / divisor); // upto here we know number
is prime or not.
if (remainder_dump != 0)
{
++i;
printf("%d.\t\t\t\t", number_to_be_checked); //print if prime.
};
number_to_be_checked = number_to_be_checked + 1;
} while (number_to_be_checked <= upper_limit);
printf("\n pi(x) = %d \n", i);
//printf("pi function value is %f.", (i - 1) / (log(i - 1)));
float app;
app = upper_limit / (log(upper_limit));
float plot_value;
plot_value = (i) / app;
printf(" BETA FUNCTION VALUE ~ %f", plot_value);
return 0;
}