Subtract numbers using arrays - C++

Question

I want to calculate the difference between two numbers (let's say v and n, so v-n) using arrays (don't ask why I have to do so). The arrays for each number are made in the following way:

Their capacity is the number of digits of the greatest number between v and n (=q in the code)
vArray[i] = ith digit of v except leading zeros to fill the whole array
nArray[i] = - ith digit of n except leading zeros to fill the whole array

For example, choose v = 10 and n = 2 then,

vArray = [1,0]
nArray = [0,-2]

So I wrote this code to calculate the sum array that will be equal to the digits of the difference (sum = [0,9] for the example above):

long r = 0;
for (int i = q-1 ; i > -1; i--){
    sum[i] = vArray[i] + nArray[i];
    if (sum[i] < 0){
        r = floor(sum[i]/10);
        sum[i-1] -= r;
        sum[i] = sum[i]+10;
    }else{
        r = 0;
    }

    NSLog(@"%li",sum[i]);
}

The problem is that sum array isn't equal to what it should be. For the same example, sum = [1,8] What is the problem in the code?

note : vArray and nArray are properly generated.

EDIT : A few examples and expected results

    v =  |    n =   |  vArray =   |     nArray=    |    sum=
    25   |    9     |    [2,5]    |      [0,9]     |    [1,6]
    105  |    10    |   [1,0,5]   |     [0,1,0]    |   [0,9,5]
   1956  |   132    |  [1,9,5,6]  |    [0,1,3,2]   |  [1,8,2,4]
  369375 |   6593   |[3,6,9,3,7,5]|  [0,0,6,5,9,3] |[3,6,2,7,8,2]

Could it be integer division: `sum[i]/10`? Remember, 1/2 == 0. — Thomas Matthews, Feb 21 '13 at 15:38
I don't understand your post. Is each array element one digit or more than one? What does `q` represent? — Thomas Matthews, Feb 21 '13 at 15:43
"don't ask why I have to do so" - Sorry, but that's pretty much a prequesite at that point. — Etienne de Martel, Feb 21 '13 at 15:45
`q = ceil(log10(max(v,n)))` or if you prefer, the number of digit of the largest between the v and n. Each arrays contains the associated number's digits and if needed, leading zeros. — moray95, Feb 21 '13 at 15:49
Reminds me of http://www.mikedoesweb.com/wp-content/uploads/2012/05/20091116-so-large.gif — R. Martinho Fernandes, Feb 21 '13 at 15:50
I know it's a stupid way to compute but I need this. Please help me. — moray95, Feb 21 '13 at 15:52
Please edit your post to provide 2 more examples using known values and the expected results. — Thomas Matthews, Feb 21 '13 at 17:34

score 2 · Accepted Answer · edited May 23 '17 at 11:59

2

I believe I understand the data structure, as you are using a Big Integer representation.

Given the number: 1234

Your V array is: [1, 2, 3, 4].

To add all the digits (a.k.a. sum), which I don't see why you want to do this, is:

int digit_sum = 0;
for (int i = 0;  i < 4; i++)
{
    digit_sum += v[i];
}

To convert the representation into "normal", try this:

int value = 0;
for (int i = 0; i < 4; ++i)
{
  value = (value * 10) + v[i];
}

To perform a subtraction, you will have to perform the steps as if you doing this by hand. Also, you would need a second number too.

Edit 1: link to big number subtraction
This might help:
Big Number Subtraction in C
C++ Large Number Arithmetic

edited May 23 '17 at 11:59

Community

1
1

answered Feb 21 '13 at 16:13

Thomas Matthews

56,849
17
98
154

Well what you are saying is true but doesn't have nothing with my code I don't to convert vArray or nArray into a integer. – moray95 Feb 21 '13 at 16:18
Please give some examples of known numbers and the expected results. – Thomas Matthews Feb 21 '13 at 17:33
Your link is broken man! And I'll edit the question for examples – moray95 Feb 21 '13 at 18:17
@moray95: Fixed links. Search SO for "C++ big number subtraction" or "C++ big number arithmetic". – Thomas Matthews Feb 21 '13 at 18:26

Subtract numbers using arrays - C++

1 Answers1

Linked