Revision dad96fb5233638adc567dd7f776c6a8952f2038d authored by Andreas Noack on 30 December 2017, 05:49:46 UTC, committed by GitHub on 30 December 2017, 05:49:46 UTC
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traits.jl
# This file is a part of Julia. License is MIT: https://julialang.org/license

## numeric/object traits
# trait for objects that have an ordering
abstract type TypeOrder end
struct HasOrder <: TypeOrder end
struct Unordered <: TypeOrder end

TypeOrder(instance) = TypeOrder(typeof(instance))
TypeOrder(::Type{<:Real}) = HasOrder()
TypeOrder(::Type{<:Any}) = Unordered()

# trait for objects that support arithmetic
abstract type TypeArithmetic end
struct ArithmeticRounds <: TypeArithmetic end     # least significant bits can be lost
struct ArithmeticOverflows <: TypeArithmetic end  #  most significant bits can be lost
struct ArithmeticUnknown <: TypeArithmetic end

TypeArithmetic(instance) = TypeArithmetic(typeof(instance))
TypeArithmetic(::Type{<:AbstractFloat}) = ArithmeticRounds()
TypeArithmetic(::Type{<:Integer}) = ArithmeticOverflows()
TypeArithmetic(::Type{<:Any}) = ArithmeticUnknown()

# trait for objects that support ranges with regular step
"""
    TypeRangeStep(instance)
    TypeRangeStep(T::Type)

Indicate whether an instance or a type supports constructing a range with
a perfectly regular step or not. A regular step means that
[`step`](@ref) will always be exactly equal to the difference between two
subsequent elements in a range, i.e. for a range `r::AbstractRange{T}`:
```julia
all(diff(r) .== step(r))
```

When a type `T` always leads to ranges with regular steps, it should
define the following method:
```julia
Base.TypeRangeStep(::Type{<:AbstractRange{<:T}}) = Base.RangeStepRegular()
```
This will allow [`hash`](@ref) to use an O(1) algorithm for `AbstractRange{T}`
objects instead of the default O(N) algorithm (with N the length of the range).

In some cases, whether the step will be regular depends not only on the
element type `T`, but also on the type of the step `S`. In that case, more
specific methods should be defined:
```julia
Base.TypeRangeStep(::Type{<:OrdinalRange{<:T, <:S}}) = Base.RangeStepRegular()
```

By default, all range types are assumed to be `RangeStepIrregular`, except
ranges with an element type which is a subtype of `Integer`.
"""
abstract type TypeRangeStep end
struct RangeStepRegular   <: TypeRangeStep end # range with regular step
struct RangeStepIrregular <: TypeRangeStep end # range with rounding error

TypeRangeStep(instance) = TypeRangeStep(typeof(instance))
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