Make pop!(::Set{A}, ::B) return an A, not B
#52017
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Previously, `pop!(::Set, x)` returned `x`, not the element in the set. This matters if multiple different elements are equal and hash to the same. Before: ```julia julia> pop!(Set([1]), 0x01) 0x01 ``` Now: ```julia julia> pop!(Set([1]), 0x01) 1 ```
vtjnash
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Nov 4, 2023
vtjnash
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Nov 4, 2023
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CI failures appear unrelated. |
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Would pkgeval worth it? |
oscardssmith
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Nov 10, 2023
Minor optimization to compute index in `Dict` only once. This PR should not be merged before #52017. **Master** ``` julia 126.417 μs (1 allocation: 16 bytes) 147.812 μs (1 allocation: 16 bytes) ``` **PR** ``` julia 86.494 μs (1 allocation: 16 bytes) 156.912 μs (1 allocation: 16 bytes) ``` <details> <summary><b><u>Testing code</u></b></summary> ``` julia using BenchmarkTools function PR_pop!(s::Set, x, default) dict = s.dict index = Base.ht_keyindex(dict, x) if index > 0 @inbounds key = dict.keys[index] Base._delete!(dict, index) return key else return default end end N = 10000 x = collect(1:N) x_negative = collect(-N:-1) function pop_all(s, x) for v in x pop!(s, v, -1) end end function pop_all_PR(s, x) for v in x PR_pop!(s, v, -1) end end # Master @Btime pop_all(s, x) setup=(s=Set(x)) @Btime pop_all(s, x_negative) setup=(s=Set(x)) # PR @Btime pop_all_PR(s, x) setup=(s=Set(x)) @Btime pop_all_PR(s, x_negative) setup=(s=Set(x)) ``` </details>
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Previously,
pop!(::Set, x)returnedx, not the element in the set.This matters if multiple different elements are equal and hash to the same.I believe this might have been an oversight. It was not tested for, and it makes most sense to me that
pop!returns the element actually in the set.Before:
Now: