How can I take specific user inputs in an array?

Question

I am coding a program that converts a binary number into decimal number by doubling (link to wikihow article).

If the user input is something other than 1 or 0, then its not a binary number, under that circumstance I want the loop to "break" and say something like:

"Oops! Binary numbers have only 1 or 0".

If not "then" the loop should continue.

That is I want to code something like

for(int digits = 0; digits != digitsINbinNum; ++digits){
      if(a condition that checks if user input is anything else than 1 or 0){
         coût << ""Oops! Binary numbers have only 1 or 0" << endl; 
         break;
          }else{
        cin >> binArray[digits];/*<-----------Here's the part where I am trying to do that*/
      }
    }

Refer to the code given below for more info:

#include <iostream>
#include <iterator>

using namespace std;

int main(){
    int digitsINbinNum;
    cout << "If you don't mind. Please enter the number of digits in your binary number: ";
    cin >> digitsINbinNum;
    int binArray[digitsINbinNum];

    cout << "Enter the binary number: ";
    for(int digits = 0; digits != digitsINbinNum; ++digits){
        cin >> binArray[digits];/*<-----------Here's the part where I am trying to do that*/
    }

/*using the doubling method as found in wikihow.com*/
    int total = 0;
    for(int posiOFdigit = 0; posiOFdigit != sizeof(binNum[noOFdigits]); posiOFdigit++){
        total = total * 2 + binNum[posiOFdigit];
    }

    /*Printing the number*/
    cout << "Decimal form of ";
    for(int n = 0; n != noOFdigits; n++){
        cout << binNum[n];
    }
    cout << " is " << total;
    return 0;
}

Answer 1

The logic for converting a binary number into decimal number by the doubling method can be referred from the given link in the question.

Modifying the given code to keep it as close as possible to the question's reference code.

Note: As ISO C++ forbids variable length array, I am changing int binArray[digits] to int *binArray = (int *)malloc(sizeof(int) * digitsINbinNum);. This modification makes it an integer pointer and it gets the memory of required size allocated at runtime.

#include <iostream>

using namespace std;

int main(){
    int digitsINbinNum,
        /* variable to keep decimal conversion of given binary */
        decimal_val = 0;
    bool is_binary = true;
    cout << "If you don't mind. Please enter the number of digits in your binary number: ";
    cin >> digitsINbinNum;
    /*
     ISO C++ forbids variable length array,
     making it int pointer and allocating dynamic memory
    */
    int *binArray = (int *)malloc(sizeof(int) * digitsINbinNum);
    if (binArray == NULL)
    {
        cout << "Memory allocation failure" << endl;
        exit -1;
    }
    cout << "Enter the binary number: ";
    for(int digits = 0; digits != digitsINbinNum; ++digits){
        cin >> binArray[digits];
        /*<-----------Here's the part where I am trying to do that*/
        /* doubling method logic for conversion of given binary to decimal */
        if ((binArray[digits] == 0) ||
            (binArray[digits] == 1))
        {
            decimal_val = (decimal_val * 2) + binArray[digits];
        }
        else /* not a binary number */
        {
            is_binary = false;
            cout << "Oops! Binary numbers have only 1 or 0" << endl;
            break;
        }
    }

    /* if conversion is successful: print result */
    if (is_binary)
    {   
        cout << "Decimal Value for given binary is: " << decimal_val << endl;
    }
    if (binArray)
    {
        free(binArray);
    }
    return 0;
}

Answer 2

You don't need an array for this. Here is a simple solution:

#include <iostream>

int main(){
    int digitsINbinNum;
    std::cout << "If you don't mind. Please enter the number of digits in your binary number: ";
    std::cin >> digitsINbinNum;

    std::cout << "Enter the binary number: ";
    int ret = 0;
    for(int digits = 0; digits != digitsINbinNum; ++digits) {
        int bin;
        std::cin >> bin;
        if (bin == 1 || bin == 0) {
            ret = 2 * ret + bin;
        } else {
            std::cout << "Oops! Binary numbers have only 1 or 0" << std::endl;
            return -1;
        }
    }
    std::cout << ret << std::endl;

    return 0;
}