# Long Math Operation result is 0, In Python isn't

Hi everyone
Well, I’m new in this in general and just testing the REPL with random numbers.
Why this is happening and what I’ve to learn previous to declare operations like this one?

In Julia (1.7.0):

``````n = 123 * 123123567 * 5675675 * 86786 * 456456 * 78678 * 354345345 * 567567 * 3545 * 678678 * 234^23423 * 238^345354
--
0
``````

To confirm if the result is truly 0 I’ve tested in Python:
In Python (3.10.1)

``````n
--
30332308162763280714160902682533172328953144277024172702616
``````

I’ve already test with Real, Int8, Int16, Int64, Float64 before each operation and every print in Julia gives me 0.

The problem is that integers overflow, and thus:

``````julia> 234^23423
0

julia> 235^23423
-5572363361863899709
``````

1 Like
``````234^23423 * 238^345354
``````

At first glance, you can tell that these numbers do not fit into a `64` bit integer, and hence we’ll only get the lower 64 bits (which will all be zero, as `true_answer % (2^64) == 0`, which you can tell from the fact that both numbers are even and both exponents are greater than 64).
Multiplying the other numbers by `0` will of course result in an answer of `0`.

Convert the numbers to `BigInt` first.

``````n = big"123" * 123123567 * 5675675 * 86786 * 456456 * 78678 * 354345345 * 567567 * 3545 * 678678 * big(234)^23423 * big"238"^34535
``````

This is probably enough to be correct (multipications should go left to right, so making the first `big` should keep all intermediates big, and then we also need the two exponents).
The number is very large.
Far larger than the number you said Python gave.
I’d post it, but it is several times over the character limit.

6 Likes

Use `BigInt`, e.g by: `big"234"^23423`.
Also note, that in python the `^` operator is the `xor` operator, while in julia it is the exponentiation operator.
So unless you actually want `xor`s, the result you got in python is not the one you expect.
(In python, the exponentiation operator is `**`)

5 Likes

Ah, that’s why the number was so small in Python.
You can also use `⊻` (type via `/xor`) in Julia.

Using `xor`, I now get:

``````julia> n = big"123" * 123123567 * 5675675 * 86786 * 456456 * 78678 * 354345345 * 567567 * 3545 * 678678 * 234 ⊻ 23423 * 238 ⊻ 34535
30332308162763280714160902682533172328953144277024172521589
``````
4 Likes

Thank You @trahflow @Elrod

At first I’ve tested the `BigInt` but just at the beginning. Now I write:

n = BigInt(123123 * 3453453 * 564456456 * 234234 * 7567567 * 5345345 * 567567 * (big(5523)^238489) * (big(837218)^2342340))

And I think the result is right (Really big number)

When you write:

``````n = BigInt(123123 * 3453453 * 564456456 * 234234 * 7567567 * 5345345 * 567567 * (big(5523)^238489) * (big(837218)^2342340))
``````

Note that

``````123123 * 3453453 * 564456456 * 234234 * 7567567 * 5345345 * 567567
``````

will be calculated with all of them as `Int`s, which will overflow.

``````123123 * 3453453 * 564456456 * 234234 * 7567567 * 5345345 * 567567 * (big(5523)^238489) * (big(837218)^2342340)
``````

will be a `BigInt`, therefore calling `BigInt` on the above result will not do anything.
Hence:

``````n = BigInt(123123) * 3453453 * 564456456 * 234234 * 7567567 * 5345345 * 567567 * (big(5523)^238489) * (big(837218)^2342340)
``````

is what you want.
Making the first of these a `BigInt` will keep the accumulation in `BigInt` as we move from left to right.
And, we also still need the exponents to be `BigInt` as well, hence I left the `big` on them.

8 Likes

Two packages you may find interesting/useful:

• CheckedArithmetic.jl provides a macro which checks for over/underflows in calculations.

• SafeREPL.jl changes all number literals in the REPL to a different type, BigInts by default, so that you don’t get under/overflow. This will make most calculations slower, but is useful if you care more about correctness than speed.

2 Likes