Why does Julia have multiple ways to define functions?

[xiaodai]

@hatmatrix

The docs present at least three ways: the standard form, “compact” form, anonymous form, etc. Then there are the do blocks…

there are about 20 ways Meet some mind-blowing ways to define functions in Julia: a tour of the many ways to create functions in Julia | by ZJ Dai | Medium

lol

[hatmatrix]

It’s cool that the language gives you flexibility to do all these things (like the method of overloading) but this is in ridiculous territory!

[xiaodai]

actually i feel uneasy about this. I would prefer fewer ways to define functions

[lmiq]

@sadish-d

Another example: implicit multiplication: a=[1,2]; a'a gives you 5. But [1,2]'[1,2] gives 2.

I honestly still find this confusing. What is happening to for that to result be 2?

julia> ([1,2]')[1,2]
2

julia> ([1,2]') * [1,2]
5

Oh, I got it. Yeah, that’s quite confusing.

[abraemer]

@lmiq

I honestly still find this confusing. What is happening to for that to result be 2?

It is an indexing operation actually It just happens to work out with the specific numbers in this example.
So ([1,2]') just gives the “row vector” [1;;2] and the following [1,2] indexes it at position 1,2 so first row, second column which is… 2

Consider as a sanity check:

julia> ([1,2]')[1,3]
ERROR: BoundsError: attempt to access 1×2 adjoint(::Vector{Int64}) with eltype Int64 at index [1, 3]
Stacktrace:
...

So in the end, this confusion is due to a clash of syntax between indexing and array literals that just happens to work out accidentally with the numbers in this example.