Hi I was wondering if it is possible to apply a short circuit evaluation to 2 bitvectors. Let’s say I want to implement the or operator on 2 bitvectors, I can do that with .|. But this operator is not short circuited. I tried doing it with .|| but this gave me an error. Is there any way around this?
This has been added, it’s included since Julia 1.7.:
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You can do this, but it’s not a good idea. Short-circuit evaluation is there to avoid doing expensive calculations. Evaluating a bitwise or on two bitvectors is as cheap as it gets. Using a short-circuit version will introduce a branch, which is much more expensive, and prevent simd-vectorization.
Example:
julia> @btime (a .| b) setup=(a=bitrand(1024); b=bitrand(1024));
39.919 ns (2 allocations: 224 bytes)
julia> @btime (a .|| b) setup=(a=bitrand(1024); b=bitrand(1024));
1.030 μs (3 allocations: 4.41 KiB)
As you can see, the short-circuit version is 25x slower.
If, on the other hand, you have some expensive operations, short-circuiting pays off:
julia> using Primes
julia> @btime isodd.(a) .| isprime.(b) setup=(a=rand(10^3:10^4, 1024);b=rand(10^3:10^4, 1024));
52.200 μs (127 allocations: 6.34 KiB)
julia> @btime isodd.(a) .|| isprime.(b) setup=(a=rand(10^3:10^4, 1024);b=rand(10^3:10^4, 1024));
27.400 μs (57 allocations: 5.25 KiB)
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In my case it is quite relevant to have a short-circuit evaluation. This is because the second expression only gives an output if the first expression was false. A MWE is shown below:
using Distributions
a = rand(100)
b = rand(100)
c = (b .>= a) .|| (rand.(Bernoulli.(b./a)))
If the expression wasn’t short-circuited it would throw off an error in my case.
But thanks anyways!
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