Fix bugs for x^Val and add tests

[kshyatt]

The Float16 method was trying to fall back to something that doesn't exist...

[tkelman]

cc @stevengj ref 6c6bbba

[stevengj]

This used to work, but it was invalidated by #20889. It should have been changed to:

@inline literal_pow{p}(::typeof(^), x::Float16, ::Type{Val{p}}) = Float16(literal_pow(^,Float32(x),Val{p}))

We shouldn't be defining ^ methods for Val at all any more.

[tkelman]

so these are also incorrect now then?

base/gmp.jl:447:^{p}(x::BigInt, ::Type{Val{p}}) = x^p
base/irrationals.jl:215:^{p}(::Irrational{:e}, ::Type{Val{p}}) = exp(p)
base/mpfr.jl:508:^{p}(x::BigFloat, ::Type{Val{p}}) = x^p

[stevengj]

(We could also just delete this method and use the fallback, since performance probably isn't really an issue for this method: the fallback Float16(Float32(x)^Float32(p)) should be fast enough.)

[stevengj]

@tkelman, yes, I commented on those below — those methods can now just be deleted. (The fallback literal_pow will now do the right thing for those types.)

[kshyatt]

Cheers, glad we pinged you on this! I'll update the PR.

[kshyatt]

I've removed the methods I added and put in the literal_pow one and get:

Test threw an exception of type MethodError
  Expression: x ^ Val{1} === x
  MethodError: no method matching ^(::Float32, ::Type{Val{1}})
  Closest candidates are:
    ^(::Float32, !Matched::Float32) at math.jl:697
    ^(::Float32, !Matched::Integer) at math.jl:699
    ^(!Matched::BigInt, ::Type{Val{p}}) where p at gmp.jl:447

Do I need to add a line for the Float32 and Float64 types similar to Float16?

[stevengj]

The problem is with the tests; I added a comment explaining why.

[stevengj]

There are also ^{p}(x::BigFloat, ::Type{Val{p}}) and ^{p}(::Irrational{:e}, ::Type{Val{p}}) methods that should be deleted, since they will now be handled by the literal_pow fallback.

[stevengj]

This test (and the similar tests below) should be removed, since there are (or should be) no Val methods for ^.

To test literal_pow, just @test f^1 === Float16(2f0), since f^1 gets converted into a literal_pow call.

[kshyatt]

Thanks!