What is the meaning of the first {T} in this function declaration? Is it a return type? function zca{T}(o::Whiten{T}) y = .... Newbie, coming from numpy.

More concretely, the stuff that appears within those {} brackets in function definition syntax f{…}(…) define the names that are used as TypeVars in the argument list that follows. TypeVars are kinda like wildcards; they try to match against the arguments that you pass. Those names are then bound t…

Welcome! From Parametric Types in the manual (although the question pertains more to Parametric Methods ): it can be any type at all (or a value of any bits type) It means that function zca can have a family of methods, like: function zca{Int32}(o::Whiten{Int32}) function zca{Float64}(o::Whiten…

[image] akis: That advises the compiler to produce separate code for each method (when a different parameter gets actually used), resulting in faster code. Note that this is the case even if f is defined as: function zca(o::Whiten) I’m not exactly sure what the advantage is of the {T} notat…

[image] akis: That advises the compiler to produce separate code for each method (when a different parameter gets actually used), resulting in faster code. That’s not true. Julia will always specialize functions concrete types. In fact, [image] akis: function zca{Int32}(o::Whiten{Int32}) …

And just a quick aside (since you’re referring to my code)… in this specific case the brackets were unnecessary, but the function just above uses the type T in the body of the code to ensure type stability of the constant: function whitened_pca{T}(o::Whiten{T}) y = normal_pca(o) y[:] = y ./…

How about function whitened_pca(o::Whiten) y = normal_pca(o) y[:] = y ./ sqrt.(o.e .+ 1//100_000_000) y end ? This is better in the case T==BigFloat, but may be worse performance-wise.

Meaning of {T}

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Meaning of {T}

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