需要帮助在Prime Emirp C ++中发现错误

Question

Hey I've been trying to figure out the error in this code Im supposed to ask the user for a positive integer then pint out the first emirps 5 on each line... I'm just flat out stuck at this point ..thanks

  #include <iostream>
   using namespace std;
  int isPrime(int value); //Prototyle for "prime number function"
  int reverse (int value2); //Prototype for "emirp function"

  int main()
 {

//Ask the user for a positive number

cout << "Please enter a positive number: ";
int n;
cin >> n;

//Reject negative value input
if ( n < 1)
{
    cout << "INVALID NUMBER \n";
}
else

    //Calculate all emirps up to 'n'.
    for (int test = 0; test < n; test++)
    {
        int number = 2;

        if (isPrime(number))
        {
            cout << "\n" << reverse(number) << "\t\t\t";
        }
    }

return 0;
  }

  int isPrime(int value)
 {
//If value is prime, the remainder (count) will be zero twice--for 1 and itself.
int divisor = 1;
int count = 0;
int prime = 0;
if (value % divisor == 0)
{
    count++;
    ++divisor;
}
if ((count = 2))
{
    int prime = value; //store prime value in new variable
}

return prime;
}


int reverse(int value2)
{
//reverse the number
value2*=10;
value2 = value2 %10;
value2/=10;

//same procedure as prime function
int divisor2 = 1;
int count2 = 0;
int emirp = 0;
if (value2 % divisor2 == 0)
{//if

        count2++;
        ++divisor2;
    }
    if ((count2 = 2))
    {
        int emirp = value2;
    }
return emirp;

system ("pause");

Answer 1

Please take some time to describe your problem properly. My psychic powers tell me that the user enters a number and the program then will print all prime number up to this number with the digits reversed. (Some punster chose to call a prime number with reversed digits an Emirp.)

Hey I've been trying to figure out the error in this code ...

Hey, there's not one single error here. The code is strewn with errors!

  #include <iostream>
   using namespace std;
  int isPrime(int value);
  int reverse (int value2);

  int main()
 {

cout << "Please enter a positive number: ";
int n;
cin >> n;

if ( n < 1)
{
    cout << "INVALID NUMBER \n";
}
else

    //Calculate all emirps up to 'n'.
    for (int test = 0; test < n; test++)   ## No need to test 0 or 1 for primality
    {
        int number = 2;

        if (isPrime(number))   ## You always test whether 2 is a prime here
        {
            cout << "\n" << reverse(number) << "\t\t\t";
        }
    }

return 0;
  }

  int isPrime(int value)
 {
//If value is prime, the remainder (count) will be zero twice--for 1 and itself.
int divisor = 1;
int count = 0;
int prime = 0;       ## (A)
if (value % divisor == 0)
{
    count++;         ## This statelment will be executed at most once
                     ## You should count divisors in a loop
    ++divisor;       ## Here, divisor is incremented, but never used again
                     ## Also, if this were inside a loop, you should increment
                     ## the divisor unconsitionally. The condition affects
                     ## only the count.
}
if ((count = 2))     ## This will set count to 2 and always evaluate to true
{
    int prime = value;   ## This variable will shadow the "prime" variable 
                         ## decralered at (A). This variable will cease to exist
                         ## as soon as the block closes, i.e. immedietely.
                         ## You just want "prime = 1", without the "int", which
                         ## declares a new variable.
}

return prime;       ## This prime is tze one declared at (A) and will be 0.
}


int reverse(int value2)
{
value2*=10;
value2 = value2 %10;    ## The remainder of a division by 10 of a number that
                        ## you have just multiplied by 10 is 0, rendering pretty
                        ## much the rest of the function useless ...
value2/=10;

int divisor2 = 1;       ## ... which doesn't hurt, because the rest of the
                        ## function was useless to start with. Reversing the
                        ## digits of a number isn't at all like finding a prime.
int count2 = 0;
int emirp = 0;
if (value2 % divisor2 == 0)
{

        count2++;
        ++divisor2;
    }
    if ((count2 = 2))
    {
        int emirp = value2;
    }
return emirp;

system ("pause");      ## First, this statement comes directly after a terurn, 
                       ## so it will never be reached. Second, if anywhere, it
                       ## should go into the main routine. You don't want the 
                       ## user to press a key before printing each number.

}

Please, learn:

• how to step through a prigram with a debugger to learn how te variables change and what a program actually does;
• how loops and scope blocks (in curly braces) work;
• when to declare new variables and whan to use existing variables; (You'll want the latter more often than you think);
• to organise your program better, it will help you to spot logical errors.

As to your problems at hand: There are plenty of resources for testing for prime numbers and reversing the digits of a number on SO, which shouldn't be hard to find.