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[SR-4279] Eager prefix(while:) on sequence(...).lazy instance #46862

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palimondo mannequin opened this issue Mar 17, 2017 · 3 comments
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[SR-4279] Eager prefix(while:) on sequence(...).lazy instance #46862

palimondo mannequin opened this issue Mar 17, 2017 · 3 comments
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@palimondo
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bug A deviation from expected or documented behavior. Also: expected but undesirable behavior. compiler The Swift compiler in itself type checker Area → compiler: Semantic analysis

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@palimondo
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Mannequin

palimondo mannequin commented Mar 17, 2017

Previous ID SR-4279
Radar rdar://problem/31194055
Original Reporter @palimondo
Type Bug
Status Resolved
Resolution Done

Attachment: Download

Environment

Xcode Version 8.3 beta 4 (8W143q)

Additional Detail from JIRA
Votes 0
Component/s Compiler
Labels Bug, TypeChecker
Assignee @palimondo
Priority Medium

md5: 07b8a4858b3678887ebe1acf699ebb5b

relates to:

  • SR-1856 prefix() chained after sorted() and before map() confuses type inference in 3.0

Issue Description:

Executing following test case never terminates:

extension Sequence {
    func _length() -> Int {
        return reduce(0, {$0.0 + 1})
    }
}

func lengthOfBoundedInfiniteSequence() -> Int {
    return sequence(first: 0, next: {$0}).lazy.prefix(while: {$0 < 2}).prefix(16)._length()
}

lengthOfBoundedInfiniteSequence()

If you start splitting the chain of sequence operations to 1 and 2 subexpressions, compiler gives error: ambiguous use of 'prefix(while: )’.

func lengthOfBoundedInfiniteSequence1() -> Int {
    let a = sequence(first: 0, next: {$0})
    return a.lazy.prefix(while: {$0 < 2}).prefix(16)._length()
}
lengthOfBoundedInfiniteSequence1()

func lengthOfBoundedInfiniteSequence2() -> Int {
    let a = sequence(first: 0, next: {$0})
    let b = a.lazy
    return b.prefix(while: {$0 < 2}).prefix(16)._length()
}
lengthOfBoundedInfiniteSequence2()

When you split it to 3 subexpressions, all of them type inferred, suddenly all works as it should:

func lengthOfBoundedInfiniteSequence3() -> Int {
    let a = sequence(first: 0, next: {$0})
    let b = a.lazy
    let c = b.prefix(while: {$0 < 2})
    return c.prefix(16)._length()
}
lengthOfBoundedInfiniteSequence3() // returns 16

This looks like it’s related to SR-4256? (cc @rudkx)

Note: You can ignore the attached main.swift - that was before I minimized the test case.
@palimondo
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palimondo mannequin commented Mar 17, 2017

This splitting also works correctly:

func lengthOfBoundedInfiniteSequence4() -> Int {
    let a = sequence(first: 0, next: {$0}).lazy.prefix(while: {$0 < 2})
    return a.prefix(16)._length()
}
lengthOfBoundedInfiniteSequence4()

@airspeedswift
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@swift-ci create

@xedin
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xedin commented Nov 18, 2017

@palimondo I think this has been fixed on master but signature of some stdlib functions has changed, can you please verify using nightly and resolve if everything is ok, otherwise please assign to me!

@swift-ci swift-ci transferred this issue from apple/swift-issues Apr 25, 2022
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@palimondo palimondo

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bug A deviation from expected or documented behavior. Also: expected but undesirable behavior. compiler The Swift compiler in itself type checker Area → compiler: Semantic analysis
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