Aliasing of slices
How to check whether two slices are backed up by the same array?
For example:
a := int{1, 2, 3}
b := a[0:1]
c := a[2:3]
alias(b, c) == true
How should alias
look like?
arrays pointers go memory slice
add a comment |
How to check whether two slices are backed up by the same array?
For example:
a := int{1, 2, 3}
b := a[0:1]
c := a[2:3]
alias(b, c) == true
How should alias
look like?
arrays pointers go memory slice
1
You must compare the addresses of their elements. Check if the ranges overlap.
– Volker
Nov 13 at 11:54
add a comment |
How to check whether two slices are backed up by the same array?
For example:
a := int{1, 2, 3}
b := a[0:1]
c := a[2:3]
alias(b, c) == true
How should alias
look like?
arrays pointers go memory slice
How to check whether two slices are backed up by the same array?
For example:
a := int{1, 2, 3}
b := a[0:1]
c := a[2:3]
alias(b, c) == true
How should alias
look like?
arrays pointers go memory slice
arrays pointers go memory slice
edited Nov 13 at 12:18
icza
162k24314354
162k24314354
asked Nov 13 at 11:47
Ecir Hana
3,822104184
3,822104184
1
You must compare the addresses of their elements. Check if the ranges overlap.
– Volker
Nov 13 at 11:54
add a comment |
1
You must compare the addresses of their elements. Check if the ranges overlap.
– Volker
Nov 13 at 11:54
1
1
You must compare the addresses of their elements. Check if the ranges overlap.
– Volker
Nov 13 at 11:54
You must compare the addresses of their elements. Check if the ranges overlap.
– Volker
Nov 13 at 11:54
add a comment |
2 Answers
2
active
oldest
votes
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 {
if len(a) == 0 || len(b) == 0 {
return false
}
amin := reflect.ValueOf(&a[0]).Pointer()
amax := reflect.ValueOf(&a[len(a)-1]).Pointer()
bmin := reflect.ValueOf(&b[0]).Pointer()
bmax := reflect.ValueOf(&b[len(b)-1]).Pointer()
return !(amax < bmin || amin > bmax)
}
Testing it:
a := int{0, 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.
b, c and d are all backed by the same array, sooverlap(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 ofx[cap(x)-1]
: play.golang.org/p/ULB3RJhz07i
– Peter
Nov 13 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 at 14:47
@Peter I've misunderstood the question. Editing now.
– icza
Nov 13 at 14:59
@EcirHana Misunderstood, edited the question to answer that.
– icza
Nov 13 at 15:05
add a comment |
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).
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 {
if len(a) == 0 || len(b) == 0 {
return false
}
amin := reflect.ValueOf(&a[0]).Pointer()
amax := reflect.ValueOf(&a[len(a)-1]).Pointer()
bmin := reflect.ValueOf(&b[0]).Pointer()
bmax := reflect.ValueOf(&b[len(b)-1]).Pointer()
return !(amax < bmin || amin > bmax)
}
Testing it:
a := int{0, 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.
b, c and d are all backed by the same array, sooverlap(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 ofx[cap(x)-1]
: play.golang.org/p/ULB3RJhz07i
– Peter
Nov 13 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 at 14:47
@Peter I've misunderstood the question. Editing now.
– icza
Nov 13 at 14:59
@EcirHana Misunderstood, edited the question to answer that.
– icza
Nov 13 at 15:05
add a comment |
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 {
if len(a) == 0 || len(b) == 0 {
return false
}
amin := reflect.ValueOf(&a[0]).Pointer()
amax := reflect.ValueOf(&a[len(a)-1]).Pointer()
bmin := reflect.ValueOf(&b[0]).Pointer()
bmax := reflect.ValueOf(&b[len(b)-1]).Pointer()
return !(amax < bmin || amin > bmax)
}
Testing it:
a := int{0, 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.
b, c and d are all backed by the same array, sooverlap(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 ofx[cap(x)-1]
: play.golang.org/p/ULB3RJhz07i
– Peter
Nov 13 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 at 14:47
@Peter I've misunderstood the question. Editing now.
– icza
Nov 13 at 14:59
@EcirHana Misunderstood, edited the question to answer that.
– icza
Nov 13 at 15:05
add a comment |
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 {
if len(a) == 0 || len(b) == 0 {
return false
}
amin := reflect.ValueOf(&a[0]).Pointer()
amax := reflect.ValueOf(&a[len(a)-1]).Pointer()
bmin := reflect.ValueOf(&b[0]).Pointer()
bmax := reflect.ValueOf(&b[len(b)-1]).Pointer()
return !(amax < bmin || amin > bmax)
}
Testing it:
a := int{0, 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.
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 {
if len(a) == 0 || len(b) == 0 {
return false
}
amin := reflect.ValueOf(&a[0]).Pointer()
amax := reflect.ValueOf(&a[len(a)-1]).Pointer()
bmin := reflect.ValueOf(&b[0]).Pointer()
bmax := reflect.ValueOf(&b[len(b)-1]).Pointer()
return !(amax < bmin || amin > bmax)
}
Testing it:
a := int{0, 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.
edited Nov 13 at 15:01
answered Nov 13 at 12:16
icza
162k24314354
162k24314354
b, c and d are all backed by the same array, sooverlap(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 ofx[cap(x)-1]
: play.golang.org/p/ULB3RJhz07i
– Peter
Nov 13 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 at 14:47
@Peter I've misunderstood the question. Editing now.
– icza
Nov 13 at 14:59
@EcirHana Misunderstood, edited the question to answer that.
– icza
Nov 13 at 15:05
add a comment |
b, c and d are all backed by the same array, sooverlap(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 ofx[cap(x)-1]
: play.golang.org/p/ULB3RJhz07i
– Peter
Nov 13 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 at 14:47
@Peter I've misunderstood the question. Editing now.
– icza
Nov 13 at 14:59
@EcirHana Misunderstood, edited the question to answer that.
– icza
Nov 13 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 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 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 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 at 14:47
@Peter I've misunderstood the question. Editing now.
– icza
Nov 13 at 14:59
@Peter I've misunderstood the question. Editing now.
– icza
Nov 13 at 14:59
@EcirHana Misunderstood, edited the question to answer that.
– icza
Nov 13 at 15:05
@EcirHana Misunderstood, edited the question to answer that.
– icza
Nov 13 at 15:05
add a comment |
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).
add a comment |
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).
add a comment |
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).
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).
answered Nov 13 at 14:45
Ecir Hana
3,822104184
3,822104184
add a comment |
add a comment |
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1
You must compare the addresses of their elements. Check if the ranges overlap.
– Volker
Nov 13 at 11:54