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WhateverCode in given doesn't get refreshed when entering surrounding block in Rakudo #4901

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p6rt opened this issue Dec 21, 2015 · 5 comments
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@p6rt
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@p6rt p6rt commented Dec 21, 2015

Migrated from rt.perl.org#126984 (status was 'resolved')

Searchable as RT126984$

@p6rt
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@p6rt p6rt commented Dec 21, 2015

From @masak

<masak> m​: sub foo($x) { say (* == $x)($_) given $x }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«True␤False␤»
* masak submits rakudobug

Examining the expression printed, it basically says "$x should be
numerically equal to itself" in a circuitous way. Since this is true
for any integer, I'd expect the program to print "True\nTrue\n".

Here's what's wrong​:

<masak> m​: sub foo($x) { say (* ~ $x)($_) given $x }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«11␤21␤»

That "1" there in "21" is from the first call to &foo -- it shouldn't
be there any more. But something in the WhateverCode (* ~ $x) holds
onto the old $x.

<masak> this bug brought to you by​: Refactoring™
<lucasb> so... the WhateverCode is caching the first value it's *created* with?
<masak> yes, but only if it's in a given
<masak> m​: sub foo($x) { say (* ~ $x)($x) }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«11␤22␤»

@p6rt
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@p6rt p6rt commented Sep 15, 2017

From @skids

On Mon, 21 Dec 2015 13​:33​:51 -0800, masak wrote​:

<masak> m​: sub foo($x) { say (* == $x)($_) given $x }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«True␤False␤»
* masak submits rakudobug

Examining the expression printed, it basically says "$x should be
numerically equal to itself" in a circuitous way. Since this is true
for any integer, I'd expect the program to print "True\nTrue\n".

Here's what's wrong​:

<masak> m​: sub foo($x) { say (* ~ $x)($_) given $x }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«11␤21␤»

That "1" there in "21" is from the first call to &foo -- it shouldn't
be there any more. But something in the WhateverCode (* ~ $x) holds
onto the old $x.

<masak> this bug brought to you by​: Refactoring™
<lucasb> so... the WhateverCode is caching the first value it's
*created* with?
<masak> yes, but only if it's in a given
<masak> m​: sub foo($x) { say (* ~ $x)($x) }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«11␤22␤»

Fudged tests preemptively added to S02-types/whatever.t in roast commit 4b2fef409.

@p6rt
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@p6rt p6rt commented Sep 15, 2017

The RT System itself - Status changed from 'new' to 'open'

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@p6rt p6rt commented Jan 26, 2018

From @zoffixznet

On Mon, 21 Dec 2015 13​:33​:51 -0800, masak wrote​:

<masak> m​: sub foo($x) { say (* == $x)($_) given $x }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«True␤False␤»
* masak submits rakudobug

Examining the expression printed, it basically says "$x should be
numerically equal to itself" in a circuitous way. Since this is true
for any integer, I'd expect the program to print "True\nTrue\n".

Here's what's wrong​:

<masak> m​: sub foo($x) { say (* ~ $x)($_) given $x }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«11␤21␤»

That "1" there in "21" is from the first call to &foo -- it shouldn't
be there any more. But something in the WhateverCode (* ~ $x) holds
onto the old $x.

<masak> this bug brought to you by​: Refactoring™
<lucasb> so... the WhateverCode is caching the first value it's
*created* with?
<masak> yes, but only if it's in a given
<masak> m​: sub foo($x) { say (* ~ $x)($x) }; foo(1); foo(2)
<camelia> rakudo-moar cfb1f3​: OUTPUT«11␤22␤»

Thank you for the report. This is now fixed.

Fix​: rakudo/rakudo@1ee89b5
Test​: Raku/roast@2f29987

@p6rt
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@p6rt p6rt commented Jan 26, 2018

@zoffixznet - Status changed from 'open' to 'resolved'

@p6rt p6rt closed this Jan 26, 2018
@p6rt p6rt added the Bug label Jan 5, 2020
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