Consider the following simplified code for Julia 0.6: using BenchmarkTools @noinline g(x) = x[1] function f1(M, ::Type{Val{N}}) where N s = 0 for j = 1:M x = ntuple(i -> i + j, Val{N}) s += g(x) end return s end @generated function f2(M, ::Type{Val{N}}) where N …

Performant way to create ntuple?

GunnarFarneback September 8, 2017, 8:53pm 3

Thanks. I’ve read that issue before but apparently not actively enough to make the connection. This solves my non-simplified problem as well. Great help.

1 Like

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Performant way to create ntuple?

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