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fix power_by_squaring: use promote instead of type inference #55634

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Oct 8, 2024
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fix power_by_squaring: use promote instead of type inference #55634

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6 changes: 5 additions & 1 deletion base/intfuncs.jl
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Original file line number Diff line number Diff line change
Expand Up @@ -298,7 +298,11 @@ function invmod(n::T) where {T<:BitInteger}
end

# ^ for any x supporting *
to_power_type(x) = convert(Base._return_type(*, Tuple{typeof(x), typeof(x)}), x)
function to_power_type(x::Number)
T = promote_type(typeof(x), typeof(one(x)), typeof(x*x))
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Suggested change
T = promote_type(typeof(x), typeof(one(x)), typeof(x*x))
T = promote_type(typeof(x), typeof(x*x))

It just occured to me that a multiplicative identity doesn't necessarily exist:

https://en.wikipedia.org/wiki/Rng_(algebra)

Are such structures supported?

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I guess it'd be OK to require a type without a multiplicative identity to implement ^ themselves?

convert(T, x)
end
to_power_type(x) = oftype(x*x, x)
@noinline throw_domerr_powbysq(::Any, p) = throw(DomainError(p, LazyString(
"Cannot raise an integer x to a negative power ", p, ".",
"\nConvert input to float.")))
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22 changes: 22 additions & 0 deletions test/math.jl
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Expand Up @@ -1498,6 +1498,28 @@ end
n = Int64(1024 / log2(E))
@test E^n == Inf
@test E^float(n) == Inf

# #55633
struct Issue55633_1 <: Number end
struct Issue55633_3 <: Number end
struct Issue55633_9 <: Number end
Base.one(::Issue55633_3) = Issue55633_1()
Base.:(*)(::Issue55633_3, ::Issue55633_3) = Issue55633_9()
Base.promote_rule(::Type{Issue55633_1}, ::Type{Issue55633_3}) = Int
Base.promote_rule(::Type{Issue55633_3}, ::Type{Issue55633_9}) = Int
Base.promote_rule(::Type{Issue55633_1}, ::Type{Issue55633_9}) = Int
Base.promote_rule(::Type{Issue55633_1}, ::Type{Int}) = Int
Base.promote_rule(::Type{Issue55633_3}, ::Type{Int}) = Int
Base.promote_rule(::Type{Issue55633_9}, ::Type{Int}) = Int
Base.convert(::Type{Int}, ::Issue55633_1) = 1
Base.convert(::Type{Int}, ::Issue55633_3) = 3
Base.convert(::Type{Int}, ::Issue55633_9) = 9
for x ∈ (im, pi, Issue55633_3())
p = promote(one(x), x, x*x)
for y ∈ 0:2
@test all((t -> ===(t...)), zip(x^y, p[y + 1]))
end
end
end
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# Test that sqrt behaves correctly and doesn't exhibit fp80 double rounding.
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