Aliasing of slices










2















How to check whether two slices are backed up by the same array?



For example:



a := int1, 2, 3
b := a[0:1]
c := a[2:3]

alias(b, c) == true


How should alias look like?










share|improve this question



















  • 1





    You must compare the addresses of their elements. Check if the ranges overlap.

    – Volker
    Nov 13 '18 at 11:54















2















How to check whether two slices are backed up by the same array?



For example:



a := int1, 2, 3
b := a[0:1]
c := a[2:3]

alias(b, c) == true


How should alias look like?










share|improve this question



















  • 1





    You must compare the addresses of their elements. Check if the ranges overlap.

    – Volker
    Nov 13 '18 at 11:54













2












2








2


1






How to check whether two slices are backed up by the same array?



For example:



a := int1, 2, 3
b := a[0:1]
c := a[2:3]

alias(b, c) == true


How should alias look like?










share|improve this question
















How to check whether two slices are backed up by the same array?



For example:



a := int1, 2, 3
b := a[0:1]
c := a[2:3]

alias(b, c) == true


How should alias look like?







arrays pointers go memory slice






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 13 '18 at 12:18









icza

175k25353383




175k25353383










asked Nov 13 '18 at 11:47









Ecir HanaEcir Hana

3,975104287




3,975104287







  • 1





    You must compare the addresses of their elements. Check if the ranges overlap.

    – Volker
    Nov 13 '18 at 11:54












  • 1





    You must compare the addresses of their elements. Check if the ranges overlap.

    – Volker
    Nov 13 '18 at 11:54







1




1





You must compare the addresses of their elements. Check if the ranges overlap.

– Volker
Nov 13 '18 at 11:54





You must compare the addresses of their elements. Check if the ranges overlap.

– Volker
Nov 13 '18 at 11:54












2 Answers
2






active

oldest

votes


















3














In general you can't tell if the backing array is shared between 2 slices, because using a full slice expression, one might control the capacity of the resulting slice, and then there will be no overlap even when checking the capacity.



As an example, if you have a backing array with 10 elements, a slice may be created that only contains the first 2 elements, and its capacity might be 2. And another slice may be create that only holds its last 2 elements, its capacity again being 2.



See this example:



a := [10]int

x := a[0:2:2]
y := a[8:10:10]

fmt.Println("len(x) = ", len(x), ", cap(x) = ", cap(x))
fmt.Println("len(y) = ", len(y), ", cap(y) = ", cap(y))


The above will print that both lengths and capcities of x and y are 2. They obviously have the same backing array, but you won't have any means to tell that.




Edit: I've misunderstood the question, and the following describes how to tell if (elements of) 2 slices overlap.



There is no language support for this, but since slices have a contiguous section of some backing array, we can check if the address range of their elements overlap.



Unfortunately pointers are not ordered in the sense that we can't apply the < and > operators on them (there are pointers in Go, but there is no pointer arithmetic). And checking if all the addresses of the elements of the first slice matches any from the second, that's not feasible.



But we can obtain a pointer value (an address) as a type of uintptr using the reflect package, more specifically the Value.Pointer() method (or we could also do that using package unsafe, but reflect is "safer"), and uintptr values are integers, they are ordered, so we can compare them.



So what we can do is obtain the addresses of the first and last elements of the slices, and by comparing them, we can tell if they overlap.



Here's a simple implementation:



func overlap(a, b int) bool 


Testing it:



a := int0, 1, 2, 3
b := a[0:2]
c := a[2:4]
d := a[0:3]

fmt.Println(overlap(a, b)) // true
fmt.Println(overlap(b, c)) // false
fmt.Println(overlap(c, d)) // true


Try it on the Go Playground.






share|improve this answer

























  • b, c and d are all backed by the same array, so overlap(b, c) should return true as well; as shown in the question (and the name is misleading). Unless I'm missing something, it should be enough to compare the addresses of x[cap(x)-1]: play.golang.org/p/ULB3RJhz07i

    – Peter
    Nov 13 '18 at 14:37












  • Thanks but the question is whether the backing array is the same (and not whether the slices overlap).

    – Ecir Hana
    Nov 13 '18 at 14:47











  • @Peter I've misunderstood the question. Editing now.

    – icza
    Nov 13 '18 at 14:59











  • @EcirHana Misunderstood, edited the question to answer that.

    – icza
    Nov 13 '18 at 15:05


















0














Found one way of this here. The idea is that while I don't think there's a way of finding the beginning of the backing array, ptr + cap of a slice should[*] point to the end of it. So then one compares the last pointer for equality, like:



func alias(x, y nat) bool 
return cap(x) > 0 && cap(y) > 0 && &x[0:cap(x)][cap(x)-1] == &y[0:cap(y)][cap(y)-1]



[*] The code includes the following note:




Note: alias assumes that the capacity of underlying arrays is never changed for nat values; i.e. that there are no 3-operand slice expressions in this code (or worse, reflect-based operations to the same effect).







share|improve this answer






















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    2 Answers
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    2 Answers
    2






    active

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    oldest

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    active

    oldest

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    3














    In general you can't tell if the backing array is shared between 2 slices, because using a full slice expression, one might control the capacity of the resulting slice, and then there will be no overlap even when checking the capacity.



    As an example, if you have a backing array with 10 elements, a slice may be created that only contains the first 2 elements, and its capacity might be 2. And another slice may be create that only holds its last 2 elements, its capacity again being 2.



    See this example:



    a := [10]int

    x := a[0:2:2]
    y := a[8:10:10]

    fmt.Println("len(x) = ", len(x), ", cap(x) = ", cap(x))
    fmt.Println("len(y) = ", len(y), ", cap(y) = ", cap(y))


    The above will print that both lengths and capcities of x and y are 2. They obviously have the same backing array, but you won't have any means to tell that.




    Edit: I've misunderstood the question, and the following describes how to tell if (elements of) 2 slices overlap.



    There is no language support for this, but since slices have a contiguous section of some backing array, we can check if the address range of their elements overlap.



    Unfortunately pointers are not ordered in the sense that we can't apply the < and > operators on them (there are pointers in Go, but there is no pointer arithmetic). And checking if all the addresses of the elements of the first slice matches any from the second, that's not feasible.



    But we can obtain a pointer value (an address) as a type of uintptr using the reflect package, more specifically the Value.Pointer() method (or we could also do that using package unsafe, but reflect is "safer"), and uintptr values are integers, they are ordered, so we can compare them.



    So what we can do is obtain the addresses of the first and last elements of the slices, and by comparing them, we can tell if they overlap.



    Here's a simple implementation:



    func overlap(a, b int) bool 


    Testing it:



    a := int0, 1, 2, 3
    b := a[0:2]
    c := a[2:4]
    d := a[0:3]

    fmt.Println(overlap(a, b)) // true
    fmt.Println(overlap(b, c)) // false
    fmt.Println(overlap(c, d)) // true


    Try it on the Go Playground.






    share|improve this answer

























    • b, c and d are all backed by the same array, so overlap(b, c) should return true as well; as shown in the question (and the name is misleading). Unless I'm missing something, it should be enough to compare the addresses of x[cap(x)-1]: play.golang.org/p/ULB3RJhz07i

      – Peter
      Nov 13 '18 at 14:37












    • Thanks but the question is whether the backing array is the same (and not whether the slices overlap).

      – Ecir Hana
      Nov 13 '18 at 14:47











    • @Peter I've misunderstood the question. Editing now.

      – icza
      Nov 13 '18 at 14:59











    • @EcirHana Misunderstood, edited the question to answer that.

      – icza
      Nov 13 '18 at 15:05















    3














    In general you can't tell if the backing array is shared between 2 slices, because using a full slice expression, one might control the capacity of the resulting slice, and then there will be no overlap even when checking the capacity.



    As an example, if you have a backing array with 10 elements, a slice may be created that only contains the first 2 elements, and its capacity might be 2. And another slice may be create that only holds its last 2 elements, its capacity again being 2.



    See this example:



    a := [10]int

    x := a[0:2:2]
    y := a[8:10:10]

    fmt.Println("len(x) = ", len(x), ", cap(x) = ", cap(x))
    fmt.Println("len(y) = ", len(y), ", cap(y) = ", cap(y))


    The above will print that both lengths and capcities of x and y are 2. They obviously have the same backing array, but you won't have any means to tell that.




    Edit: I've misunderstood the question, and the following describes how to tell if (elements of) 2 slices overlap.



    There is no language support for this, but since slices have a contiguous section of some backing array, we can check if the address range of their elements overlap.



    Unfortunately pointers are not ordered in the sense that we can't apply the < and > operators on them (there are pointers in Go, but there is no pointer arithmetic). And checking if all the addresses of the elements of the first slice matches any from the second, that's not feasible.



    But we can obtain a pointer value (an address) as a type of uintptr using the reflect package, more specifically the Value.Pointer() method (or we could also do that using package unsafe, but reflect is "safer"), and uintptr values are integers, they are ordered, so we can compare them.



    So what we can do is obtain the addresses of the first and last elements of the slices, and by comparing them, we can tell if they overlap.



    Here's a simple implementation:



    func overlap(a, b int) bool 


    Testing it:



    a := int0, 1, 2, 3
    b := a[0:2]
    c := a[2:4]
    d := a[0:3]

    fmt.Println(overlap(a, b)) // true
    fmt.Println(overlap(b, c)) // false
    fmt.Println(overlap(c, d)) // true


    Try it on the Go Playground.






    share|improve this answer

























    • b, c and d are all backed by the same array, so overlap(b, c) should return true as well; as shown in the question (and the name is misleading). Unless I'm missing something, it should be enough to compare the addresses of x[cap(x)-1]: play.golang.org/p/ULB3RJhz07i

      – Peter
      Nov 13 '18 at 14:37












    • Thanks but the question is whether the backing array is the same (and not whether the slices overlap).

      – Ecir Hana
      Nov 13 '18 at 14:47











    • @Peter I've misunderstood the question. Editing now.

      – icza
      Nov 13 '18 at 14:59











    • @EcirHana Misunderstood, edited the question to answer that.

      – icza
      Nov 13 '18 at 15:05













    3












    3








    3







    In general you can't tell if the backing array is shared between 2 slices, because using a full slice expression, one might control the capacity of the resulting slice, and then there will be no overlap even when checking the capacity.



    As an example, if you have a backing array with 10 elements, a slice may be created that only contains the first 2 elements, and its capacity might be 2. And another slice may be create that only holds its last 2 elements, its capacity again being 2.



    See this example:



    a := [10]int

    x := a[0:2:2]
    y := a[8:10:10]

    fmt.Println("len(x) = ", len(x), ", cap(x) = ", cap(x))
    fmt.Println("len(y) = ", len(y), ", cap(y) = ", cap(y))


    The above will print that both lengths and capcities of x and y are 2. They obviously have the same backing array, but you won't have any means to tell that.




    Edit: I've misunderstood the question, and the following describes how to tell if (elements of) 2 slices overlap.



    There is no language support for this, but since slices have a contiguous section of some backing array, we can check if the address range of their elements overlap.



    Unfortunately pointers are not ordered in the sense that we can't apply the < and > operators on them (there are pointers in Go, but there is no pointer arithmetic). And checking if all the addresses of the elements of the first slice matches any from the second, that's not feasible.



    But we can obtain a pointer value (an address) as a type of uintptr using the reflect package, more specifically the Value.Pointer() method (or we could also do that using package unsafe, but reflect is "safer"), and uintptr values are integers, they are ordered, so we can compare them.



    So what we can do is obtain the addresses of the first and last elements of the slices, and by comparing them, we can tell if they overlap.



    Here's a simple implementation:



    func overlap(a, b int) bool 


    Testing it:



    a := int0, 1, 2, 3
    b := a[0:2]
    c := a[2:4]
    d := a[0:3]

    fmt.Println(overlap(a, b)) // true
    fmt.Println(overlap(b, c)) // false
    fmt.Println(overlap(c, d)) // true


    Try it on the Go Playground.






    share|improve this answer















    In general you can't tell if the backing array is shared between 2 slices, because using a full slice expression, one might control the capacity of the resulting slice, and then there will be no overlap even when checking the capacity.



    As an example, if you have a backing array with 10 elements, a slice may be created that only contains the first 2 elements, and its capacity might be 2. And another slice may be create that only holds its last 2 elements, its capacity again being 2.



    See this example:



    a := [10]int

    x := a[0:2:2]
    y := a[8:10:10]

    fmt.Println("len(x) = ", len(x), ", cap(x) = ", cap(x))
    fmt.Println("len(y) = ", len(y), ", cap(y) = ", cap(y))


    The above will print that both lengths and capcities of x and y are 2. They obviously have the same backing array, but you won't have any means to tell that.




    Edit: I've misunderstood the question, and the following describes how to tell if (elements of) 2 slices overlap.



    There is no language support for this, but since slices have a contiguous section of some backing array, we can check if the address range of their elements overlap.



    Unfortunately pointers are not ordered in the sense that we can't apply the < and > operators on them (there are pointers in Go, but there is no pointer arithmetic). And checking if all the addresses of the elements of the first slice matches any from the second, that's not feasible.



    But we can obtain a pointer value (an address) as a type of uintptr using the reflect package, more specifically the Value.Pointer() method (or we could also do that using package unsafe, but reflect is "safer"), and uintptr values are integers, they are ordered, so we can compare them.



    So what we can do is obtain the addresses of the first and last elements of the slices, and by comparing them, we can tell if they overlap.



    Here's a simple implementation:



    func overlap(a, b int) bool 


    Testing it:



    a := int0, 1, 2, 3
    b := a[0:2]
    c := a[2:4]
    d := a[0:3]

    fmt.Println(overlap(a, b)) // true
    fmt.Println(overlap(b, c)) // false
    fmt.Println(overlap(c, d)) // true


    Try it on the Go Playground.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 13 '18 at 15:01

























    answered Nov 13 '18 at 12:16









    iczaicza

    175k25353383




    175k25353383












    • b, c and d are all backed by the same array, so overlap(b, c) should return true as well; as shown in the question (and the name is misleading). Unless I'm missing something, it should be enough to compare the addresses of x[cap(x)-1]: play.golang.org/p/ULB3RJhz07i

      – Peter
      Nov 13 '18 at 14:37












    • Thanks but the question is whether the backing array is the same (and not whether the slices overlap).

      – Ecir Hana
      Nov 13 '18 at 14:47











    • @Peter I've misunderstood the question. Editing now.

      – icza
      Nov 13 '18 at 14:59











    • @EcirHana Misunderstood, edited the question to answer that.

      – icza
      Nov 13 '18 at 15:05

















    • b, c and d are all backed by the same array, so overlap(b, c) should return true as well; as shown in the question (and the name is misleading). Unless I'm missing something, it should be enough to compare the addresses of x[cap(x)-1]: play.golang.org/p/ULB3RJhz07i

      – Peter
      Nov 13 '18 at 14:37












    • Thanks but the question is whether the backing array is the same (and not whether the slices overlap).

      – Ecir Hana
      Nov 13 '18 at 14:47











    • @Peter I've misunderstood the question. Editing now.

      – icza
      Nov 13 '18 at 14:59











    • @EcirHana Misunderstood, edited the question to answer that.

      – icza
      Nov 13 '18 at 15:05
















    b, c and d are all backed by the same array, so overlap(b, c) should return true as well; as shown in the question (and the name is misleading). Unless I'm missing something, it should be enough to compare the addresses of x[cap(x)-1]: play.golang.org/p/ULB3RJhz07i

    – Peter
    Nov 13 '18 at 14:37






    b, c and d are all backed by the same array, so overlap(b, c) should return true as well; as shown in the question (and the name is misleading). Unless I'm missing something, it should be enough to compare the addresses of x[cap(x)-1]: play.golang.org/p/ULB3RJhz07i

    – Peter
    Nov 13 '18 at 14:37














    Thanks but the question is whether the backing array is the same (and not whether the slices overlap).

    – Ecir Hana
    Nov 13 '18 at 14:47





    Thanks but the question is whether the backing array is the same (and not whether the slices overlap).

    – Ecir Hana
    Nov 13 '18 at 14:47













    @Peter I've misunderstood the question. Editing now.

    – icza
    Nov 13 '18 at 14:59





    @Peter I've misunderstood the question. Editing now.

    – icza
    Nov 13 '18 at 14:59













    @EcirHana Misunderstood, edited the question to answer that.

    – icza
    Nov 13 '18 at 15:05





    @EcirHana Misunderstood, edited the question to answer that.

    – icza
    Nov 13 '18 at 15:05













    0














    Found one way of this here. The idea is that while I don't think there's a way of finding the beginning of the backing array, ptr + cap of a slice should[*] point to the end of it. So then one compares the last pointer for equality, like:



    func alias(x, y nat) bool 
    return cap(x) > 0 && cap(y) > 0 && &x[0:cap(x)][cap(x)-1] == &y[0:cap(y)][cap(y)-1]



    [*] The code includes the following note:




    Note: alias assumes that the capacity of underlying arrays is never changed for nat values; i.e. that there are no 3-operand slice expressions in this code (or worse, reflect-based operations to the same effect).







    share|improve this answer



























      0














      Found one way of this here. The idea is that while I don't think there's a way of finding the beginning of the backing array, ptr + cap of a slice should[*] point to the end of it. So then one compares the last pointer for equality, like:



      func alias(x, y nat) bool 
      return cap(x) > 0 && cap(y) > 0 && &x[0:cap(x)][cap(x)-1] == &y[0:cap(y)][cap(y)-1]



      [*] The code includes the following note:




      Note: alias assumes that the capacity of underlying arrays is never changed for nat values; i.e. that there are no 3-operand slice expressions in this code (or worse, reflect-based operations to the same effect).







      share|improve this answer

























        0












        0








        0







        Found one way of this here. The idea is that while I don't think there's a way of finding the beginning of the backing array, ptr + cap of a slice should[*] point to the end of it. So then one compares the last pointer for equality, like:



        func alias(x, y nat) bool 
        return cap(x) > 0 && cap(y) > 0 && &x[0:cap(x)][cap(x)-1] == &y[0:cap(y)][cap(y)-1]



        [*] The code includes the following note:




        Note: alias assumes that the capacity of underlying arrays is never changed for nat values; i.e. that there are no 3-operand slice expressions in this code (or worse, reflect-based operations to the same effect).







        share|improve this answer













        Found one way of this here. The idea is that while I don't think there's a way of finding the beginning of the backing array, ptr + cap of a slice should[*] point to the end of it. So then one compares the last pointer for equality, like:



        func alias(x, y nat) bool 
        return cap(x) > 0 && cap(y) > 0 && &x[0:cap(x)][cap(x)-1] == &y[0:cap(y)][cap(y)-1]



        [*] The code includes the following note:




        Note: alias assumes that the capacity of underlying arrays is never changed for nat values; i.e. that there are no 3-operand slice expressions in this code (or worse, reflect-based operations to the same effect).








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 13 '18 at 14:45









        Ecir HanaEcir Hana

        3,975104287




        3,975104287



























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            𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

            Crossroads (UK TV series)

            ữḛḳṊẴ ẋ,Ẩṙ,ỹḛẪẠứụỿṞṦ,Ṉẍừ,ứ Ị,Ḵ,ṏ ṇỪḎḰṰọửḊ ṾḨḮữẑỶṑỗḮṣṉẃ Ữẩụ,ṓ,ḹẕḪḫỞṿḭ ỒṱṨẁṋṜ ḅẈ ṉ ứṀḱṑỒḵ,ḏ,ḊḖỹẊ Ẻḷổ,ṥ ẔḲẪụḣể Ṱ ḭỏựẶ Ồ Ṩ,ẂḿṡḾồ ỗṗṡịṞẤḵṽẃ ṸḒẄẘ,ủẞẵṦṟầṓế