How to convert negative number to base P in C++

Question

I have integer num, which can be negative or positive and also I have number P (1 < P <= 9).

What I want to do is to convert num to base-P.

I easily do it when num > 0:

int i = 0;
int A[10000];
while (num != 0)
{
    A[i] = num % p;
    num /= p;
    i++;
}

But I don't know how to achieve it when num < 0.

Answer 1

The - sign is not something special to base 10. In general if 90base{10} = 1011010base{2} then -90base{10} = -1011010base{2}. What you are doing here is an extension of two's complement. Two's complement is a method to represent both positive and negative numbers in binary without using a - and is used in computing. Just check whether number is positive or negative and put negative sign. You can use for checking Adrian's answer. But, you can also use the function code that returns -1 for negative values, 0 for zero, 1 for positive values of x

int status (int x) {
  int sign = (x > 0) - (x < 0); 
  return sign;
}

Answer 2

Check if number is positive or negative, and remember this in the code. If it's negative - multiply the number by -1 and do the same you did to a positive number. When you output the number, you should do this before:

if (sign == -1)
    cout << '-';

And then write the rest of the number.