deciding if a number is perfect or prime

问题:

the problem is : "Write a function to find out if a number is a prime or perfect number."

so far i have worked on the perfect part first and this is what i have:

#include <iostream>
using namespace std;
bool perfectNumber(int);
int main()
{
 int number;

 cout<<"Please enter number:\n";
 cin>>number;
 bool perfectNumber(number);

 return 0;
}
bool perfectNumber(int number)
{
 int i;

 int sum=0;
 for(i=1;i<=number/2;i++)
 {
  if(number%i==0)
  {
   sum+=i;
  }
 }
 if (sum==number)
  return i;
 else
  return 0;
}

HOWEVER, there seems to be errors on this code. I have looked over the book but nothing talks about this topic. i would like to get advice on how to fix this code.

thanks!

回答1:

bool perfectNumber(number);

This does not call the perfectNumber function; it declares a local variable named perfectNumber of type bool and initializes it with the value of number converted to type bool.

In order to call the perfectNumber function, you need to use something along the lines of:

bool result = perfectNumber(number);

or:

bool result(perfectNumber(number));

On another note: if you are going to read input from a stream (e.g. cin>>number), you must check to be sure that the extraction of the value from the stream succeeded. As it is now, if you typed in asdf, the extraction would fail and number would be left uninitialized. The best way to check whether an extraction succeeds is simply to test the state of the stream:

if (cin >> number) {
    bool result = perfectNumber(number);
}
else {
    // input operation failed; handle the error as appropriate
}

You can learn more about how the stream error states are set and reset in Semantics of flags on basic_ios. You should also consult a good, introductory-level C++ book for more stream-use best practices.

回答2:

void primenum(long double x) {
    bool prime = true; 
    int number2;
    number2 = (int) floor(sqrt(x));// Calculates the square-root of 'x'

    for (int i = 1; i <= x; i++) {
        for (int j = 2; j <= number2; j++) {
            if (i != j && i % j == 0) {
                prime = false;
                break;
            }
        }
        if (prime) {
            cout << " " << i << " ";
            c += 1;
        }
        prime = true;
    }
}

回答3:

 bool isPerfect(  int number){
     int i;
     int sum=0;
     for(i=1;i<number ;i++){
         if(number %i == 0){
             cout<<"  " << i ;
             sum+=i;
         }
     }

     if (sum == number){
         cout<<"\n \t\t THIS NUMBER  >>>  "<<  number <<"   IS    PERFECT \n\n";
         return i;
     }else if (sum |= number) {
               cout<<"\nThis number >>> " <<  number <<"   IS  NOT  PERFECT \n\n";
               return 0;
     }
 }

回答4:

#pragma hdrstop

#include <tchar.h>
#include <stdio.h>
#include <conio.h>
//---------------------------------------------------------------------------


bool is_prim(int nr)
{

    for (int i = 2; i < nr-1; i++) {

    if (nr%i==0) return false;

    }

    return true;

}

bool is_ptr(int nr)
{
    int sum=0;

    for (int i = 1; i < nr; i++) {

    if (nr%i==0) {
        sum=sum+i;
    }

    }

    if (sum==nr) { return true;

    }
    else return false;
}
#pragma argsused
int _tmain(int argc, _TCHAR* argv[])
{
    int numar;

    printf ("Number=");scanf("%d",&numar);
    if (is_prim(numar)==true) { printf("The number is prime");

    }
    else printf("The number is not prime");

    if (is_ptr(numar)==true) { printf(" The number is perfect");

    }
    else printf(" The number is not perfect");
    getch();
    return 0;
}