Read digits from an int and counting them - c++

Question

Alright so I've been looking though Google and on forums for hours and can't seem to understand how to solve this problem.
I need to write a program that first determines if the number entered by the user is a base 5 number (in other words, a number that only has 0s, 1s, 2s, 3s, and 4s in it). Then, I have to count how many 0s, 1s, 2s, etc are in the number and display it to the user.
I've seen people saying I should convert int to a string and then using cin.get().
I noticed that I can't use cin.get() on a string, it needs to be a char.
I can only use a while loop for this assignment, no while... do loops.
Any help is appreciated!!
Here's what I have so far, obviously with all my mistakes in it:
//----------------------------------------------
// Assignment 3
// Question 1
// File name: q1.cpp
// Written by: Shawn Rousseau (ID: 7518455)
// For COMP 218 Section EC / Winter 2015
// Concordia University, Montreal, QC
//-----------------------------------------------
// The purpose of this program is to check if the
// number entered by the user is a base of 5
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
// Declaring variables
int number;
int zeros;
int ones;
int twos;
int threes;
int fours;
bool base5;
// Get data and calculate
cin >> number;
string numberString = to_string(number);
// Determine if the number is a base 5 number
while (cin.get(numberString) == 0 || cin.get(numberString) == 1 ||
cin.get(numberString) == 2 || cin.get(numberString) == 3 ||
cin.get(numberString) == 4)
base5 = true;
// Determine the number of each digits
zeros = 0;
ones = 0;
twos = 0;
threes = 0;
fours = 0;
return 0;
}

Answer 1

Several things you need to beware of:
One way to get a specific character from a std::string is by []. e.g.
std::string myString{"abcdefg"};
char myChar = myString[4]; // myChar == 'e'
cin.get(aString) is not trying to get data from aString. It continues to get data from stdin and store in aString. Once you have get the data and put into the string, you can simply manipulate the string itself.
a short piece of code that will count number of vowel in a string. If you can understand it, there should be no problem doing your work. (Haven't compiled, probably some typos)
std::string inputString;
std::cin >> inputString;
// you said you need a while loop.
// although it is easier to with for loop and iterator...
int i = 0;
int noOfVowels = 0;
while (i < inputString.length()) {
if (inputString[i] == 'a'
|| inputString[i] == 'e'
|| inputString[i] == 'i'
|| inputString[i] == 'o'
|| inputString[i] == 'u' ) {
++noOfVowels;
}
++i;
}
std::cout << "number of vowels : " << noOfVowels << std::endl;

Answer 2

Well you can try this approach. This will solve your needs I guess.
#include <iostream>
#include <string>
#include <sstream>
#include <conio.h>
using namespace std;
int _tmain(int argc, _TCHAR* argv[])
{
int number=0;
int zeros=0;
int ones=0;
int twos=0;
int threes=0;
int fours=0;
bool valid=true;;
int counter = 0;
cout<<"Enter the number: ";
cin >> number;
stringstream out;
out << number; //int to string
string numberString = out.str();
cout<<"\n\nNumber after string conversion : "<<numberString;
cout<<"\nPrinting this just to show that the conversion was successful\n\n\n";
while (counter < numberString.length())
{
if (numberString[counter] == '0')
zeros++;
else if(numberString[counter] == '1')
ones++;
else if(numberString[counter] == '2')
twos++;
else if(numberString[counter] == '3')
threes++;
else if(numberString[counter] == '4')
fours++;
else
valid=false;
counter++;
}
if(valid==true)
{
cout<<"\nZeros : "<<zeros;
cout<<"\nOnes : "<<ones;
cout<<"\nTwos : "<<twos;
cout<<"\nThrees : "<<threes;
cout<<"\nFours : "<<fours;
}
else
cout<<"\n\nInvalid data...base of 5 rule violated";
_getch();
return 0;
}

Answer 3

If you don't want to use std::string then use characters, first loop over the input from the user until ENTER is pressed.
char ch = 0;
while ((ch = cin.get()) != '\n')
{
...
}
For each character read, check if it is a digit (std::isdigit) and if it is in the range 0..4, if not quit and give some message of not being base 5
have an array of ints to keep track of the frequency of the digits
int freq[5] = {0,0,0,0,0};
after you have checked that the character is valid subtract the ascii value from the digit and use that as index in the array, increment that:
freq[ch - '0']++;
e.g.
char ch;
int freq[5] = {0};
while ((ch = cin.get()) != '\n')
{
cout << ch;
freq[ch-'0']++;
}
for (int i = 0; i < sizeof(freq)/sizeof(freq[0]); ++i)
{
cout << static_cast<char>(48+i) << ":" << freq[i] << endl;
}

Answer 4

Here's a useful function that counts digits:
// D returns the number of times 'd' appears as a digit of n.
// May not work for n = INT_MIN.
int D(int n, int d) {
// Special case if we're counting zeros of 0.
if (n == 0 && d == 0) return 1;
if (n < 0) n = -n;
int r = 0;
while (n) {
if (n % 10 == d) r++;
n /= 10;
}
return r;
}
In your code you can use this naively to solve the problem without any further loops.
if (D(n, 5) + D(n, 6) + D(n, 7) + D(n, 8) + D(n, 9) != 0) {
cout << "the number doesn't consist only of the digits 0..4."
}
cout << "0s: " << D(n, 0) << "\n";
cout << "1s: " << D(n, 1) << "\n";
cout << "2s: " << D(n, 2) << "\n";
cout << "3s: " << D(n, 3) << "\n";
cout << "4s: " << D(n, 4) << "\n";
You could also (or should also) use loops here to reduce the redundancy.