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Fix linear indexing for 0D views of OffsetVectors #39404

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Feb 1, 2021
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Fix linear indexing for 0D views of OffsetVectors #39404

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d54b4ef
Fix linear indexing for views of OffsetVectors
jishnub Jan 26, 2021
f600830
update comments
mbauman Jan 26, 2021
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6 changes: 3 additions & 3 deletions base/subarray.jl
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Original file line number Diff line number Diff line change
Expand Up @@ -380,19 +380,19 @@ compute_offset1(parent::AbstractVector, stride1::Integer, I::Tuple{AbstractRange
# indexing uses the indices along the given dimension.
# If the result is one-dimensional and it's a range, then linear
# indexing might be offset if the index itself is offset
# Otherwise linear indexing always starts with 1.
# Otherwise linear indexing always matches the parent.
compute_offset1(parent, stride1::Integer, I::Tuple) =
(@_inline_meta; compute_offset1(parent, stride1, find_extended_dims(1, I...), find_extended_inds(I...), I))
compute_offset1(parent, stride1::Integer, dims::Tuple{Int}, inds::Tuple{Slice}, I::Tuple) =
(@_inline_meta; compute_linindex(parent, I) - stride1*first(axes(parent, dims[1]))) # index-preserving case
compute_offset1(parent, stride1::Integer, dims, inds::Tuple{AbstractRange}, I::Tuple) =
(@_inline_meta; compute_linindex(parent, I) - stride1*first(axes1(inds[1]))) # potentially index-offsetting case
compute_offset1(parent, stride1::Integer, dims, inds, I::Tuple) =
(@_inline_meta; compute_linindex(parent, I) - stride1) # linear indexing starts with 1
(@_inline_meta; compute_linindex(parent, I) - stride1)
function compute_linindex(parent, I::NTuple{N,Any}) where N
@_inline_meta
IP = fill_to_length(axes(parent), OneTo(1), Val(N))
compute_linindex(1, 1, IP, I)
compute_linindex(first(LinearIndices(parent)), 1, IP, I)
end
function compute_linindex(f, s, IP::Tuple, I::Tuple{ScalarIndex, Vararg{Any}})
@_inline_meta
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9 changes: 9 additions & 0 deletions test/offsetarray.jl
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Expand Up @@ -110,6 +110,15 @@ let a1 = [11,12,13], a2 = [1 2; 3 4]
@test_throws BoundsError i2[1:2:5]
end

# issue #37274
let a = 1:3
oa = OffsetArray(a, 0:2)
b = @view oa[0]
@test b[] == b[1] == b[1,1] == 1
@test_throws BoundsError b[0]
@test_throws BoundsError b[2]
end

# logical indexing
@test A[A .> 2] == [3,4]
@test_throws BoundsError h[trues(2)]
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