Prime number C++ program - c++

Question

I am not sure whether I should ask here or programmers but I have been trying to work out why this program wont work and although I have found some bugs, it still returns "x is not a prime number", even when it is.
#include <iostream>
using namespace std;
bool primetest(int a) {
int i;
//Halve the user input to find where to stop dividing to (it will remove decimal point as it is an integer)
int b = a / 2;
//Loop through, for each division to test if it has a factor (it starts at 2, as 1 will always divide)
for (i = 2; i < b; i++) {
//If the user input has no remainder then it cannot be a prime and the loop can stop (break)
if (a % i == 0) {
return(0);
break;
}
//Other wise if the user input does have a remainder and is the last of the loop, return true (it is a prime)
else if ((a % i != 0) && (i == a -1)) {
return (1);
break;
}
}
}
int main(void) {
int user;
cout << "Enter a number to test if it is a prime or not: ";
cin >> user;
if (primetest(user)) {
cout << user << " is a prime number.";
}
else {
cout << user<< " is not a prime number.";
}
cout << "\n\nPress enter to exit...";
getchar();
getchar();
return 0;
}
Sorry if this is too localised (in which case could you suggest where I should ask such specific questions?)
I should add that I am VERY new to C++ (and programming in general)
This was simply intended to be a test of functions and controls.

Answer 1

i can never be equal to a - 1 - you're only going up to b - 1. b being a/2, that's never going to cause a match.
That means your loop ending condition that would return 1 is never true.
In the case of a prime number, you run off the end of the loop. That causes undefined behaviour, since you don't have a return statement there. Clang gave a warning, without any special flags:
example.cpp:22:1: warning: control may reach end of non-void function
[-Wreturn-type]
}
^
1 warning generated.
If your compiler didn't warn you, you need to turn on some more warning flags. For example, adding -Wall gives a warning when using GCC:
example.cpp: In function ‘bool primetest(int)’:
example.cpp:22: warning: control reaches end of non-void function
Overall, your prime-checking loop is much more complicated than it needs to be. Assuming you only care about values of a greater than or equal to 2:
bool primetest(int a)
{
int b = sqrt(a); // only need to test up to the square root of the input
for (int i = 2; i <= b; i++)
{
if (a % i == 0)
return false;
}
// if the loop completed, a is prime
return true;
}
If you want to handle all int values, you can just add an if (a < 2) return false; at the beginning.

Answer 2

Your logic is incorrect. You are using this expression (i == a -1)) which can never be true as Carl said.
For example:-
If a = 11
b = a/2 = 5 (Fractional part truncated)
So you are running loop till i<5. So i can never be equal to a-1 as max value of i in this case will be 4 and value of a-1 will be 10

Answer 3

You can do this by just checking till square root. But below is some modification to your code to make it work.
#include <iostream>
using namespace std;
bool primetest(int a) {
int i;
//Halve the user input to find where to stop dividing to (it will remove decimal point as it is an integer)
int b = a / 2;
//Loop through, for each division to test if it has a factor (it starts at 2, as 1 will always divide)
for (i = 2; i <= b; i++) {
//If the user input has no remainder then it cannot be a prime and the loop can stop (break)
if (a % i == 0) {
return(0);
}
}
//this return invokes only when it doesn't has factor
return 1;
}
int main(void) {
int user;
cout << "Enter a number to test if it is a prime or not: ";
cin >> user;
if (primetest(user)) {
cout << user << " is a prime number.";
}
else {
cout << user<< " is not a prime number.";
}
return 0;
}

Answer 4

check this out:
//Prime Numbers generation in C++
//Using for loops and conditional structures
#include <iostream>
using namespace std;
int main()
{
int a = 2; //start from 2
long long int b = 1000; //ends at 1000
for (int i = a; i <= b; i++)
{
for (int j = 2; j <= i; j++)
{
if (!(i%j)&&(i!=j)) //Condition for not prime
{
break;
}
if (j==i) //condition for Prime Numbers
{
cout << i << endl;
}
}
}
}

Answer 5

main()
{
int i,j,x,box;
for (i=10;i<=99;i++)
{
box=0;
x=i/2;
for (j=2;j<=x;j++)
if (i%j==0) box++;
if (box==0) cout<<i<<" is a prime number";
else cout<<i<<" is a composite number";
cout<<"\n";
getch();
}
}

Answer 6

Here is the complete solution for the Finding Prime numbers till any user entered number.
#include <iostream.h>
#include <conio.h>
using namespace std;
main()
{
int num, i, countFactors;
int a;
cout << "Enter number " << endl;
cin >> a;
for (num = 1; num <= a; num++)
{
countFactors = 0;
for (i = 2; i <= num; i++)
{
//if a factor exists from 2 up to the number, count Factors
if (num % i == 0)
{
countFactors++;
}
}
//a prime number has only itself as a factor
if (countFactors == 1)
{
cout << num << ", ";
}
}
getch();
}

Answer 7

One way is to use a Sieving algorithm, such as the sieve of Eratosthenes. This is a very fast method that works exceptionally well.
bool isPrime(int number){
if(number == 2 || number == 3 | number == 5 || number == 7) return true;
return ((number % 2) && (number % 3) && (number % 5) && (number % 7));
}