Question

c ++ program that converts from any base to a decimal number

Answer 1

C++ program : (Converts from any base to a decimal number)

#include <iostream>
using namespace std;
/*
To find decimal equivalent of a number N in any base we have the formula , 
let n be the length of number then, (Here ^ represents power symbol)
Decimal Equivalent = ((base)^0 )*N[n-1] + ((base)^1) * N[n-2] + ((base)^2) * N[n-3] + ....
*/
int main()
{
    string num;
    int base;
    cout << "Please enter a number: ";
    cin >> num;
    cout << "Please enter the base of the number: ";
    cin >> base;
    
    int n = num.length();  // n stores length of number
    int power = 1;
    int result = 0, digit, flag = 1; // decimal value  is stored in result
    for(int i = n-1; i>=0; i--)      // iterate through all digits in number
    {
        if(num[i]>='0' && num[i]<='9')  // convert characters to integers
        {
            digit = (int)num[i] - '0';
        }
        else                           // convert characters to integers , Here 'A' = 10, 'B' = 11 ... 'Z' = 36
        {
            digit = (int)num[i] - 'A' + 10;
        }
        
        if(digit >=base)              // if digit is greater than base then its invalid number for the given base
        {
            flag = 0;
            cout << "You have entered a invalid number for the given base\n";
            break;
        }
        result += digit*power;       // calculate result
        power = power*base;          // multiply power with base
    }
    if(flag) // if flag is set to 1 then we get decimal equivalent for valid number for that base
    {
        cout<< "Decimal equivalent of "<<num<<" in base "<<base<<" is "<<result<<"\n";
    }

    return 0;
}

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