0

I'm doing some C++ array homework. The goals is to convert decimal to binary (include negative numbers). Here's my code, it gets the job done, but I would like to see if anything can be improved, or any better algorithm (using binary shift maybe?). 

#include <iostream> 
using namespace std;
// doi tu thap phan sang nhi phan
void decToBinary(int n, int nhiphan[])
{
    for (int i=0; i < 16; i++)
    {
        // to binary
        nhiphan[i] = n % 2;
        n = n / 2;
    }
    // inverse array
    for (int i = 0, j = 15; i < j; i++, j--)
    {
        int temp = nhiphan[i];
        nhiphan[i] = nhiphan[j];
        nhiphan[j] = temp;
    }
}
void reverse(int& a)
{
    if (a == 0)
        a++;
    else a--;
}

void outArr(const int a[], int size) {
    for (int i = 0; i < size; ++i)
        cout << a[i];
}

int main()
{
    int nhiphan[16];
    int n;
    do {
        cout << "Nhap so (-255 <= n <= 255) chuyen doi sang nhi phan (16 bit): ";
        cin >> n;
    } while (n > 255 || n < -255);
    if (n < 0) {//check negative
        n *= -1;
        decToBinary(n, nhiphan);
        for (int i = 0; i < 16; i++)// 1's complement
            reverse(nhiphan[i]);
        // +1
        if (nhiphan[15] == 0)//2's complement
            nhiphan[15] = 1;
        else 
        { 
            nhiphan[15] = 0; 
            int i = 15;
            do {
                reverse(nhiphan[i-1]);
                i--;
            } while (nhiphan[i-1] == 0);
        }
    }
    else decToBinary(n, nhiphan);
    outArr(nhiphan, 16);
    return 0;
}
Phineas
  • 159
  • 2
  • 10
  • 1
    This can certainly be improved. The binary representation can be created in the right order in the first place; instead of in the wrong order and then reversed by the second loop. Then, `reverse()` can be a single statement instead of a confusing `if` statement. Finally, the shown code assumes `sizeof(int)` to be `2`. This is not necessary true. So, at least three obvious improvements here. – Sam Varshavchik Nov 23 '19 at 16:29
  • You can use [std::bitset](https://stackoverflow.com/a/2891296/4850111). – nada Nov 23 '19 at 16:55
  • @nada They need to write out the algorithm on their own. – David G Nov 23 '19 at 16:59
  • @0x499602D2 That'd be plausible but not explicitly stated in the question. – nada Nov 23 '19 at 17:01
  • @nada It's highly plausible given the use of the word "homework". – David G Nov 23 '19 at 17:03
  • There are some algorithms that allows doing some manipulations on bit (if not already supported by compiler/hardware by an intrinsic function) like https://graphics.stanford.edu/~seander/bithacks.html#BitReverseObvious but it might not be appropriate to use that in your homework... – Phil1970 Nov 23 '19 at 17:36
  • Is there any way to lock the elements of `Nhiphan[]` to only hold 1 & 0? So when I use 1+1, it automatically returns to 0 instead of 2? – Phineas Nov 24 '19 at 12:55
  • @SamVarshavchik Can you explain more about the `sizeof(int)`, specifically in this code? – Phineas Nov 24 '19 at 17:18
  • What's to explain? On 64 bit platforms `int`s are usually 4 bytes and 32 bits. – Sam Varshavchik Nov 24 '19 at 17:56
  • You said "Finally, the shown code assumes `sizeof(int)` to be 2.". I don't quite understand it. Can you explain? – Phineas Nov 25 '19 at 00:46

0 Answers0