Strange output produced by program - c++

Question

I think that my code works. However, it outputs 01111E5, or 17B879DD, or something like that.
Can someone please tell me why.
I am aware that I set the limit of P instead of 10,001. My code is like that because I start with 3, skipping the prime number 2.
#include <iostream>
bool prime (int i)
{
bool result = true;
int isitprime = i;
for(int j = 2; j < isitprime; j++) ///prime number tester
{
if(isitprime%j == 0) result = false;
}
return result;
}
int main (void)
{
using namespace std;
int PrimeNumbers = 1;
int x = 0;
for (int i = 3 ; PrimeNumbers <=10000; i++)
{
if(prime(i))
{
int prime = i;
PrimeNumbers +=1;
}
}
cout<<prime<<endl;
system ("pause");
return 0;
}

Answer 1

cout<<prime<<endl;
prints the address of the function bool prime (int i), not the variable you declared. Just rename the function or the variable (note that you'll also have to change its scope, or move the cout inside the loop - that's if you want to print them all):
for (int i = 3 ; PrimeNumbers <=10000; i++)
{
if(prime(i))
{
cout << i << endl;
PrimeNumbers++;
}
}
Also:
for(int j = 2; j < isitprime; j++) ///prime number tester
{
if(isitprime%j == 0) result = false;
}
could be optimized, since (1) you don't need to check all numbers till isitprime, but at most to sqrt(isitprimt) and (2) you only need to check until result is false, at which point you can break out of the loop.

Answer 2

The output isn't strange at all.
cout<<prime<<endl;
You're printing the function pointer of prime here.
You were probably intending to print the variable you create here:
int prime = i;
But this is in the loop scope. In fact, if you compile with warnings enabled, your compiler should tell you that this variable is never used. Also, it's bad practice to give variables in C or C++ the same name as functions (or any other variable in a higher level scope).

Answer 3

Your loop in the main program does not stop correctly because the test variable, PrimeNumbers may not change.
Try:
for (int i = 3; i < 10000; i++)
{
//...
}
Also, because you declared the variable prime inside an if statement, it disappears after the if statement is executed:
if (prime(i))
{
int prime = i; // <-- Declare the variable before the for loop.
//...

Answer 4

The code for finding prime number from nth to 2(1 is neither prime nor composite) is written below, i havent used math.h header file , did something different and surpizingly it works pretty cool....
Code is:
#include<iostream.h>
#include<stdio.h>
#include<conio.h>
class prime
{
int a;
int i;
public:
void display();
};
void prime::display()
{
cout<<"Enter the number to see primes less than it till 2";
cin>>a;
int count=0;
for(int j=a;j>=1;j--)
{
for(int i=1;i<=j;i++)
{
if(j%i==0)
{
count++;
}
}
if(count==2)
{
cout<<"\t"<<j;
}
count=0;
}
}
void main()
{
clrscr();
prime k;
k.display();
getch();
}

Answer 5

if you want to find the prime number from 1 to n,hope this will help u.
#include <iostream>
#include <vector>
static bool _isprime (int number)
{
if(number==1)
{
return false;
}
bool flag=true;
if(number==2||number%2!=0)
{
for(int i=2;i<number;i++)
{
if(number%i==0)
{
flag=false;
}
}
}
else flag=false;
return flag;
}
int main (void)
{
using namespace std;
vector<int> primenumber;
cout<<"prime number between 1 and ？"<<endl;
int x=0;
cin>>x;
for(int i=0;i<=x;i++)
{
if(_isprime(i)==true)
{
//cout<<x<<" is a prime number"<<endl;
primenumber.push_back(i);
}
//else cout<<x<<" is not a prime number"<<endl;
}
for(int i=0;i<primenumber.size();i++)
{
cout<<primenumber[i]<<endl;
}
cout<<"the number of prime number is "<<primenumber.size()<<endl;
system("pause");
return 0;
}