How to work with Base 8 (Octal) numbers? - c++

Question

In C++, an octal number is defined by preceeding it with a 0, example:
01 = 1
010 = 8
014 = 12
So I was experimenting how working with Base 8 in c++ works, and tried adding to it with a loop, like so:
int base8Number = 00;
for (int i = 01; i < 011; i+=01)
{
base8Number += i;
cout << base8Number << '\n';
}
And apparently, C++ doesn't like working with octal numbers, the output I got is as follows:
1
3
6
10
15
21
28
36
The most obvious reason I know it's not working in Base 8, is the 28 output as a result, since the 8 number isn't used in Base 8.
So, my question: Can you work with Base 8 in c++, or is it just meant to be used as a constant, and if you can work with Base 8 in c++, how do you do it?

Answer 1

So first, let's remember that when we print numbers the way you're doing, they will be shown in decimal.
Now, let's rewrite your code without octal:
int base10Number = 0;
for (int i = 1; i < 9; i+=1)
{
base10Number += i;
cout << base10Number << '\n';
}
So let's now look at what your code is actually doing:
cout << 1 << "\n"; // 1
cout << 1 + 2 << "\n"; // 3
cout << 1 + 2 + 3 << "\n"; // 6
cout << 1 + 2 + 3 + 4 << "\n"; // 10
....
Which is what you're seeing. So no, there is no problem with how octal works in c++.
If you'd like you can use std::oct to tell std::cout to use octal printing. For example:
int main() {
std::cout << std::oct << 25 << "\n"; // Outputs: 31
}

Answer 2

Remember that "base" is a property of number representation, not the number itself. If you've got enough pegs to put one on each finger, then that is the same number of pegs regardless of whether you write 10, 012 0xA, or anything else.
Your code computes the numbers which would be shown in base 10 as 1, 3, 6, 10, 15, etc. You output them in base 10. To output them in base 8 use:
std::cout << std::oct << base8Number << std::endl;

Answer 3

There are a couple of things going on here.
Intrinsically, your computer operates in binary (base 2). When you do something like int foo = 10;, you are expressing the number 10 in decimal form because it's convenient for you to read it that way, but in the end the computer will still store it using binary, e.g. 1010.
If you were to use an octal literal (e.g. 012), then as far as the computer's concerned that's still just a 1010 binary constant - the only thing that's changed is its representation in the source code.
Finally, the last thing to realise is that the computer will by default output integers in base 10, because that's what's easy for people to read. It's still outputting the number 1010, just using the decimal representation of it.
Given all of the above, your code is entirely equivalent to doing the following, which you can verify for yourself produces the same output without using any octal:
int num = 0;
for (int i = 1; i < 9; i += 1) // Constants the same, just changed from octal representation to decimal.
{
num += i;
cout << num << '\n'; // Outputs in decimal, as with original program.
}
To get what you expect, try using the oct modifier:
int base8Number = 00;
for (int i = 01; i < 011; i+=01)
{
base8Number += i;
cout << oct << base8Number << '\n';
// ^--Here. Explicitly requests octal output.
}
This then explicitly requests that the computer output the values using octal.