count the number of even, odd, and zeros in digit

[salah]

plz need help with this code(urgent)



#include<iostream>
#include<cstdlib>
#include<ctime>
#include<iomanip>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<cassert>

using namespace std;

int digits(int, int, int);

int main()
{
int ctreven, ctrodd, ctrzero;

cout<<"Enter an integer: ";

cout<<"The number of even digits: "<<digits(ctreven)<<endl;
cout<<"The number of odd digits: "<<digits(ctrodd)<<<<endl;
cout<<"The number of zeros: "<<digits(ctrzero)<< endl;

}

int digits(int ctreven, int ctrodd, int ctrzero)
{
int digit;
for(int i=0; i<digit; i++)
{
if(digit%2==0)
ctreven++;
if(digit%2!==0)
ctrodd++;
if(digit==0)
ctrzero++;
}

return digit;
}

[AbstractionAnon]

What help do you need? Please be specific.

A few obvious problems:
line 13: You're passing arguments by value. Values in main won;t be updated.
line 17: These are uninitialized variables.
line 19: You prompt for an value, but never cin a value.
line 21-23: You call digits() passing one argument, but declared digits() to take three arguments.
line 22: duplicate output operators (<<<<)
line 29: digit is an uninitialized variable (garbage).

PLEASE USE CODE TAGS (the <> formatting button) when posting code.
It makes it easier to read your code and also easier to respond to your post.
http://www.cplusplus.com/articles/jEywvCM9/
Hint: You can edit your post, highlight your code and press the <> formatting button.

[PBachmann]

I believe this is what you are looking for. Make sure to read the comments, and tell me if you need more explanation.

#include <iostream> // <-- needed for cin, cout, endl
#include <math.h> // <-- needed for pow() and log10()

using namespace std;

// This program seperates the integer into a series of digits. The digits will be reversed, but that isn't a problem.
void seperateDigits(int integer, int digitCount, int digit[]){
    for (int i = 0; i < digitCount; i++){
        digit[i] = int(integer / pow(10., double(i))); // integer / 10^i cuts off end digit (e.g. 123 first time, 12 second time, 1 third time)
        digit[i] = int(digit[i]) % 10; // Gets the final digit (e.g. 123 becomes 3, 12 becomes 2, 1 becomes 1)
    }
}

// This program will count the even, odd, and zero digits. The correct count will be assigned to ctreven, ctrodd, and ctrzero without having
// a return value. This can be done by using a "pass by refernce" method.
void countDigits(int &ctreven, int &ctrodd, int &ctrzero, int digitCount, int digit[])
{
    for(int i = 0; i < digitCount; i++)
    {
        if(digit[i] == 0){
            ctrzero++;
        }
        else if(digit[i]%2 == 0){
            ctreven++;
        }
        else{
            ctrodd++;
        }
    }
}

int main()
{
    int integer;
    // These next three need to start at 0
    int ctreven = 0;
    int ctrodd = 0;
    int ctrzero = 0;

    cout<<"Enter an integer: ";
    cin >> integer; // This is needed to accept user's input
    cout << endl; // <-- just for formatting

    int digitCount = int(log10(integer)) + 1; // log10(integer) + 1 gives you the number of digits
    int digit[digitCount]; // This array will hold the digits

    seperateDigits(integer, digitCount, digit); // This runs the function to seperate
    countDigits(ctreven, ctrodd, ctrzero, digitCount, digit); // This runs the function to count

    // Now that the digits have been counted, the 3 variables (ctreven, ctrodd, ctrzero) have the correct values

    cout << "The number of even digits: " << ctreven << endl;
    cout << "The number of odd digits: " << ctrodd << endl;
    cout << "The number of zeros: " << ctrzero << endl;

    cout << endl << endl; // <-- just for formatting

    return 0;
}

[3voN6Up4]

@PBachmann

Make sure you use comment blocks if comments are long!

#include <iostream> // <-- needed for cin, cout, endl
#include <math.h> // <-- needed for pow() and log10()

using namespace std;

/*This program seperates the integer into a series of digits. 
The digits will be reversed, but that isn't a problem.*/
void seperateDigits(int integer, int digitCount, int digit[]){
    for (int i = 0; i < digitCount; i++){
        digit[i] = int(integer / pow(10., double(i))); 
/*integer / 10^i cuts off end digit (e.g. 123 first time,
12 second time, 1 third time)*/
        digit[i] = int(digit[i]) % 10; 
/*Gets the final digit (e.g. 123 becomes 3, 
12 becomes 2, 1 becomes 1)*/
    }
}

/*This program will count the even, odd, and zero digits. 
The correct count will be assigned to ctreven, ctrodd, and 
ctrzero without having a return value. This can be done by 
using a "pass by refernce" method.*/
void countDigits(int &ctreven, int &ctrodd, int &ctrzero, int digitCount, int digit[])
{
    for(int i = 0; i < digitCount; i++)
    {
        if(digit[i] == 0){
            ctrzero++;
        }
        else if(digit[i]%2 == 0){
            ctreven++;
        }
        else{
            ctrodd++;
        }
    }
}

int main()
{
    int integer;
    // These next three need to start at 0
    int ctreven = 0;
    int ctrodd = 0;
    int ctrzero = 0;

    cout<<"Enter an integer: ";
    cin >> integer; // This is needed to accept user's input
    cout << endl; // <-- just for formatting

    int digitCount = int(log10(integer)) + 1; // log10(integer) + 1 gives you the number of digits
    int digit[digitCount]; // This array will hold the digits

    seperateDigits(integer, digitCount, digit); // This runs the function to seperate
    countDigits(ctreven, ctrodd, ctrzero, digitCount, digit); // This runs the function to count

/*Now that the digits have been counted, the 3 variables 
(ctreven, ctrodd, ctrzero) have the correct values*/

    cout << "The number of even digits: " << ctreven << endl;
    cout << "The number of odd digits: " << ctrodd << endl;
    cout << "The number of zeros: " << ctrzero << endl;

    cout << endl << endl; // <-- just for formatting

    return 0;
}