Related
Bellow is a minimal example when wrapping the OpenAL32.dll. The foreign function alcCreateContext has the argument attrlist which takes a ptr to an array of type ALCint or nil. The issue is the array can be of different lengths depending on the amount of different flags passed in. The array should be organized as [flag, int, flag, int, ...]. How can this be accomplished in a more dynamic way allowing the inclusion of ALC_FREQUENCY for example? The array size is currently hard coded into the procedure and its nasty.
when defined(windows):
{.push cdecl, dynlib: "OpenAL32.dll", importc.}
else:
{.push importc.}
type
ALCint = cint
ALCdevice* = pointer
ALCcontext* = pointer
const
ALC_MONO_SOURCES* = 0x00001010
ALC_STEREO_SOURCES* = 0x00001011
ALC_FREQUENCY* = 0x00001007
proc alcCreateContext*(device: ALCdevice; attrlist: ptr array[0..3, ALCint]): ALCcontext
proc alcOpenDevice*(devicename: cstring): ALCdevice
const attributes = [ALC_MONO_SOURCES.ALCint, 65536.ALCint, ALC_STEREO_SOURCES.ALCint, 65536.ALCint]
discard alcOpenDevice(nil).alcCreateContext(attributes.unsafeAddr)
I experimented with openArray and other containers. Is the solution some sort of cast? This is also the workaround for getting more then 256 sounds out of OpenAL.
Answer from PMunch. Thank You.
The foreign function now wants ptr UncheckedArray[ALCint] and when passing the argument use cast[ptr UncheckedArray[ALCint]](attributes.unsafeAddr)
when defined(windows):
{.push cdecl, dynlib: "OpenAL32.dll", importc.}
else:
{.push importc.}
type
ALCint = cint
ALCdevice* = pointer
ALCcontext* = pointer
const
ALC_MONO_SOURCES* = 0x00001010
ALC_STEREO_SOURCES* = 0x00001011
ALC_FREQUENCY* = 0x00001007
proc alcCreateContext*(device: ALCdevice; attrlist: ptr UncheckedArray[ALCint]): ALCcontext
proc alcOpenDevice*(devicename: cstring): ALCdevice
const attributes = [ALC_MONO_SOURCES.ALCint, 65536.ALCint, ALC_STEREO_SOURCES.ALCint, 65536.ALCint]
discard alcOpenDevice(nil).alcCreateContext(cast[ptr UncheckedArray[ALCint]](attributes.unsafeAddr))
An array in C is simply a pointer to anywhere with one or more contiguous elements of the same type. So to pass a C array to a function you simply need to get such a pointer. Say for example you have a seq of integers then the address of the first element is a C array. Simply do mySeq[0].addr and you're good. Keep the lifecycle of the data in mind though. If Nim doesn't find any more references to the sequence then the memory will get freed. You can also manually get a pointer with create (https://nim-lang.org/docs/system.html#create%2Ctypedesc) and you can cast such pointers to ptr UncheckedArray[T] to be able to use [] on the data in Nim.
If I define a new struct as
mutable struct myStruct
data::AbstractMatrix
labels::Vector{String}
end
and I want to throw an error if the length of labels is not equal to the number of columns of data, I know that I can write a constructor that enforces this condition like
myStruct(data, labels) = length(labels) != size(data)[2] ? error("Labels incorrect length") : new(data,labels)
However, once the struct is initialized, the labels field can be set to the incorrect length:
m = myStruct(randn(2,2), ["a", "b"])
m.labels = ["a"]
Is there a way to throw an error if the labels field is ever set to length not equal to the number of columns in data?
You could use StaticArrays.jl to fix the matrix and vector's sizes to begin with:
using StaticArrays
mutable struct MatVec{R, C, RC, VT, MT}
data::MMatrix{R, C, MT, RC} # RC should be R*C
labels::MVector{C, VT}
end
but there's the downside of having to compile for every concrete type with a unique permutation of type parameters R,C,MT,VT. StaticArrays also does not scale as well as normal Arrays.
If you don't restrict dimensions in the type parameters (with all those downsides) and want to throw an error at runtime, you got good and bad news.
The good news is you can control whatever mutation happens to your type. m.labels = v would call the method setproperty!(object::myStruct, name::Symbol, v), which you can define with all the safeguards you like.
The bad news is that you can't control mutation to the fields' types. push!(m.labels, 1) mutates in the push!(a::Vector{T}, item) method. The myStruct instance itself doesn't actually change; it still points to the same Vector. If you can't guarantee that you won't do something like x = m.labels; push!(x, "whoops") , then you really do need runtime checks, like iscorrect(m::myStruct) = length(m.labels) == size(m.data)[2]
A good option is to not access the fields of your struct directly. Instead, do it using a function. Eg:
mutable struct MyStruct
data::AbstractMatrix
labels::Vector{String}
end
function modify_labels(s::MyStruct, new_labels::Vector{String})
# do all checks and modifications
end
You should check chapter 8 from "Hands-On Design Patterns and Best Practices with Julia: Proven solutions to common problems in software design for Julia 1.x"
I Go, I assumed slices were passed by reference, but this seems to work for values
but not for the array itself. For example, If I have this struct:
l := Line{
Points: []Point{
Point{3, 4},
},
}
I can define a variable, which gets passed a reference to the struct's slice
slice := l.Points
And then if I modify it, the original struct referenced by the variable
is going to reflect those modifications.
slice[0].X = 1000
fmt.Printf(
"This value %d is the same as this %d",
slice[0].X,
l.Points[0].X,
)
This differs from the behavior of arrays which, I assume, are passed by value.
So, for example, if I had defined the previous code using an array:
l := Line{
Points: [1]Point{
Point{3, 4},
},
}
arr := l.Points
arr[0].X = 1000
fmt.Println(arr.[0].X != s.Points[0].X) // equals true, original struct is untouched
Then, the l struct wouldn't have been modified.
Now, if I want to modify the slice itself I obviously cannot do this:
slice = append(slice, Point{99, 100})
Since that would only redefine the slice variable, losing the original reference.
I know I can simply do this:
l.Points = append(l.Points, Point{99, 100})
But, in some cases, it is more convenient to have another variable instead of having
to type the whole thing.
I tried this:
*slice = append(*slice, Point{99, 100})
But it doesn't work as I am trying to dereference something that apparently is not a pointer.
I finally tried this:
slice := &l.Points
*slice = append(l.Points, Point{99, 100})
And it works, but I am not sure what is happening. Why is the value of slice not overwritten? How does append works here?
Let's dispense first with a terminology issue. The Go language specification does not use the word reference the way you are using it. Go does however have pointers, and pointers are a form of reference. In addition, slices and maps are kind of special as there's some underlying data—the array underneath a slice, or the storage for a map—that may or may not already exist or be created by declaring or defining a variable whose type is slice of T or map[T1]T2 for some type T or type-pair T1 and T2.1
We can take your usage of the word reference to mean explicit pointer when talking about, e.g.:
func f1(p *int) {
// code ...
}
and the implied pointer when talking about:
func f2(m map[T1]T2) { ... }
func f3(s []T) { ... }
In f1, p really is a pointer: it thus refers to some actual int, or is nil. In f2, m refers to some underlying map, or is nil. In f3, s refers to some underlying array, or is nil.
But if you write:
l := Line{
Points: []Point{
Point{3, 4},
},
}
then you must have written:
type Line struct {
// ... maybe some fields here ...
Points []Point
// ... maybe more fields here ...
}
This Line is a struct type. It is not a slice type; it is not a map type. It contains a slice type but it is not itself one.
You now talk about passing these slices. If you pass l, you're passing the entire struct by value. It's pretty important to distinguish between that, and passing the value of l.Points. The function that receives one of these arguments must declare it with the right type.
For the most part, then, talking about references is just a red herring—a distraction from what's really going on. What we need to know is: What variables are you assigning what values, using what source code?
With all of that out of the way, let's talk about your actual code samples:
l.Points = append(l.Points, Point{99, 100})
This does just what it says:
Pass l.Points to append, which is a built-in as it is somewhat magically type-flexible (vs the rest of Go, where types are pretty rigid). It takes any value of type []T (slice of T, for any valid type T) plus one or more values of type T, and produces a new value of the same type, []T.
Assigns the result to l.Points.
When append does its work, it may:
receive nil (of the given type): in this case, it creates the underlying array, or
receive a non-nil slice: in this case, it writes into the underlying array or discards that array in favor of a new larger-capacity array as needed.2
So in all cases, the underlying array may have, in effect, just been created or replaced. It's therefore important that any other use of the same underlying array be updated appropriately. Assigning the result back to l.Points updates the—presumably one-and-only—slice variable that refers to the underlying array.
We can, however, break these assumptions:
s2 := l.Points
Now l.Points and s2 both refer to the (single) underlying array. Operations that modify that underlying array will, at least potentially, affect both s2 and l.Points.
Your second example is itself OK:
*slice = append(*slice, Point{99, 100})
but you haven't shown how slice itself was declared and/or assigned-to.
Your third example is fine as well:
slice := &l.Points
*slice = append(l.Points, Point{99, 100})
The first of these lines declares-and-initializes slice to point to l.Points. The variable slice therefore has type *[]Point. Its value—the value in slice, that is, rather than that in *slice—is the address of l.Points, which has type []Point.
The value in *slice is the value in l.Points. So you could write:
*slice = append(*slice, Point{99, 100})
here. Since *slice is just another name for l.Points, you can also write:
l.Points = append(*slice, Point{99, 100})
You only need to use *slice if there's some reason that l.Points is not available,3 but you may use *slice if that's more convenient. Reading *slice reads l.Points and updating *slice updates l.Points.
1To see what I mean by may or may not be created here, consider:
var s []int
vs:
var s = []int{42}
The first leaves s == nil while the second creates an underlying array with the capacity to hold the one int value 42, holding the one int value 42, so that s != nil.
2It's not clear to me whether there is a promise never to write on an existing slice-array whose capacity is greater than its current length, but not sufficient to hold the final result. That is, can append first append 10 objects to the existing underlying array, then discover that it needs a bigger array and expand the underlying array? The difference is observable if there are other slice values referring to the existing underlying array.
3Here, a classic example would occur if you have reason to pass l.Points or &l.Points to some existing (pre-written) function:
If you need pass l.Points—the slice value—to some existing function, that existing function cannot change the slice value, but could change the underlying array. That's probably a bad plan, so if it does do this, make sure that this is OK! If it only reads the slice and underlying array, that's a lot safer.
If you need to pass &l.Points—a value that points to the slice value—to some existing function, that existing function can change both the slice, and the underlying array.
If you're writing a new function, it's up to you to write it in whatever manner is most appropriate. If you're only going to read the slice and underlying array, you can take a value of type []Point. If you intend to update the slice in place, you should take a value of type *[]Point—pointer to slice of Point.
Append returns a new slice that may modify the original backing array of the initial slice. The original slice will still point to the original backing array, not the new one (which may or may not be in the same place in memory)
For example (playground)
slice := []int{1,2,3}
fmt.Println(len(slice))
// Output: 3
newSlice := append(slice, 4)
fmt.Println(len(newSlice))
// Output: 4
fmt.Println(len(slice))
// Output: 3
While a slice can be described as a "fat pointer to an array", it is not a pointer and therefore you can't dereference it, which is why you get an error.
By creating a pointer to a slice, and using append as you did above, you are setting the slice the pointer points to to the "new" slice returned by append.
For more information, check out Go Slice Usage And Internals
Your first attempt didn't work because slices are not pointers, they can be considered reference types. Append will modify the underlying array if it has enough capacity, otherwise it returns a new slice.
You can achieve what you want with a combination of your two attempts.
playground
l := Line{
Points: []Point{
Point{3, 4},
},
}
slice := &l.Points
for i := 0; i < 100; i++ {
*slice = append(*slice, Point{99 + i, 100 + i})
}
fmt.Println(l.Points)
I know that this might be sacrilegious, but, for me, it is useful to think of slices
as structs.
type Slice struct {
len int
cap int
Array *[n]T // Pointer to array of type T
}
Since in languages like C, the [] operator is also a dereferencing operator, we can think that every time we are accessing a slice, we are actually dereferencing the underlying array and assigning some value to it. That is:
var s []int
s[0] = 1
Might be thought of as equivalent to (in pseudo-code):
var s Slice
*s.Array[0] = 1
That is why we can say that slices are "pointers". For that reason, it can modify its underlying array like this:
myArray := [3]int{1,1,1}
mySlice := myArray[0:1]
mySlice = append(mySlice, 2, 3) // myArray == mySlice
Modifying mySlice also modifies myArray, since the slice stores a pointer to the array and, on appending, we are dereferencing that pointer.
This behavior, nonetheless, is not always like this. If we exceed the capacity of the original array, a new array is created and the original array is left untouched.
myArray := [3]int{1,1,1}
mySlice := myArray[0:1]
mySlice = append(mySlice, 2, 3, 4, 5) // myArray != mySlice
The confusion arises when we try to treat the slice itself as an actual pointer. Since we can modify an underlying array by appending to it, we are led to believe that in this case:
sliceCopy := mySlice
sliceCopy = append(sliceCopy, 6)
both slices, slice and sliceCopy are the same, but they are not. We have to explicitly pass a reference to the memory address of the slice (using the & operator) in order to modify it. That is:
sliceAddress := &mySlice
*sliceAddress = append(mySlice, 6) // or append(*sliceAddress, 6)
See also
https://forum.golangbridge.org/t/slice-pass-as-value-or-pointer/2866/4
https://blog.golang.org/go-slices-usage-and-internals
https://appliedgo.net/slices/
When the formal parameter is map, assigning a value directly to a formal parameter cannot change the actual argument, but if you add a new key and value to the formal parameter, the actual argument outside the function can also be seen. Why is that?
I don't understand the output value of the following code, and the formal parameters are different from the actual parameters.
unc main() {
t := map[int]int{
1: 1,
}
fmt.Println(unsafe.Pointer(&t))
copysss(t)
fmt.Println(t)
}
func copysss(m map[int]int) {
//pointer := unsafe.Pointer(&m)
//fmt.Println(pointer)
m = map[int]int{
1: 2,
}
}
stdout :0xc000086010
map[1:1]
func main() {
t := map[int]int{
1: 1,
}
fmt.Println(unsafe.Pointer(&t))
copysss(t)
fmt.Println(t)
}
func copysss(m map[int]int) {
//pointer := unsafe.Pointer(&m)
//fmt.Println(pointer)
m[1] = 2
}
stdout :0xc00007a010
map[1:2]
func main() {
t := map[int]int{
1: 1,
}
fmt.Println(unsafe.Pointer(&t))
copysss(t)
fmt.Println(t)
}
func copysss(m map[int]int) {
pointer := unsafe.Pointer(&m)
fmt.Println(pointer)
m[1] = 2
}
stdout:0xc00008a008
0xc00008a018
map[1:2]
I want to know if the parameter is a value or a pointer.
The parameter is both a value and a pointer.
Wait.. whut?
Yes, a map (and slices, for that matter) are types, pretty similar to what you would implement. Think of a map like this:
type map struct {
// meta information on the map
meta struct{
keyT type
valueT type
len int
}
value *hashTable // pointer to the underlying data structure
}
So in your first function, where you reassign m, you're passing a copy of the struct above (pass by value), and you're assigning a new map to it, creating a new hashtable pointer in the process. The variable in the function scope is updated, but the one you passed still holds a reference to the original map, and with it, the pointer to the original map is preserved.
In the second snippet, you're accessing the underlying hash table (a copy of the pointer, but the pointer points to the same memory). You're directly manipulating the original map, because you're just changing the contents of the memory.
So TL;DR
A map is a value, containing meta information of what the map looks like, and a pointer to the actual data stored inside. The pointer is passed by value, like anything else (same way pointers are passed by value in C/C++), but of course, dereferencing a pointer means you're changing the values in memory directly.
Careful...
Like I said, slices work pretty much in the same way:
type slice struct {
meta struct {
type T
len, cap int
}
value *array // yes, it's a pointer to an underlying array
}
The underlying array is of say, a slice of ints will be [10]int if the cap of the slice is 10, regardless of the length. A slice is managed by the go runtime, so if you exceed the capacity, a new array is allocated (twice the cap of the previous one), the existing data is copied over, and the slice value field is set to point to the new array. That's the reason why append returns the slice that you're appending to, the underlying pointer may have changed etc.. you can find more in-depth information on this.
The thing you have to be careful with is that a function like this:
func update(s []int) {
for i, v := range s {
s[i] = v*2
}
}
will behave much in the same way as the function you have were you're assigning m[1] = 2, but once you start appending, the runtime is free to move the underlying array around, and point to a new memory address. So bottom line: maps and slices have an internal pointer, which can produce side-effects, but you're better off avoiding bugs/ambiguities. Go supports multiple return values, so just return a slice if you set about changing it.
Notes:
In your attempt to figure out what a map is (reference, value, pointer...), I noticed you tried this:
pointer := unsafe.Pointer(&m)
fmt.Println(pointer)
What you're doing there, is actually printing the address of the argument variable, not any address that actually corresponds to the map itself. the argument passed to unsafe.Pointer isn't of the type map[int]int, but rather it's of type *map[int]int.
Personally, I think there's too much confusion around passing by value vs passing by . Go works exactly like C in this regard, just like C, absolutely everything is passed by value. It just so happens that this value can sometimes be a memory address (pointer).
More details (references)
Slices: usage & internals
Maps Note: there's some confusion caused by this one, as pointers, slices, and maps are referred to as *reference types*, but as explained by others, and elsewhere, this is not to be confused with C++ references
In Go, map is a reference type. This means that the map actually resides in the heap and variable is just a pointer to that.
The map is passed by copy. You can change the local copy in your function, but this will not be reflected in caller's scope.
But, since the map variable is a pointer to the unique map residing in the heap, every change can be seen by any variable that points to the same map.
This article can clarify the concept: https://www.ardanlabs.com/blog/2014/12/using-pointers-in-go.html.
In GO when I use a struct as a key for a map, there is an unicity of the keys.
For example, the following code produce a map with only one key : map[{x 1}:1]
package main
import (
"fmt"
)
type MyT struct {
A string
B int
}
func main() {
dic := make(map[MyT]int)
for i := 1; i <= 10; i++ {
dic[MyT{"x", 1}] = 1
}
fmt.Println(dic)
}
// result : map[{x 1}:1]
I Tried to do the same in Julia and I had a strange surprise :
This Julia code, similar to the GO one, produces a dictionary whith 10 keys !
type MyT
A::String
B::Int64
end
dic = Dict{MyT, Int64}()
for i in 1:10
dic[MyT("x", 1)] = 1
end
println(dic)
# Dict(MyT("x",1)=>1,MyT("x",1)=>1,MyT("x",1)=>1,MyT("x",1)=>1,MyT("x",1)=>1,MyT("x",1)=>1,MyT("x",1)=>1,MyT("x",1)=>1,MyT("x",1)=>1,MyT("x",1)=>1)
println(keys(dic))
# MyT[MyT("x",1),MyT("x",1),MyT("x",1),MyT("x",1),MyT("x",1),MyT("x",1),MyT("x",1),MyT("x",1),MyT("x",1),MyT("x",1)]
So what I did wrong ?
Thank you #DanGetz for the solution ! :
immutable MyT # or struct MyT with julia > 0.6
A::String
B::Int64
end
dic = Dict{MyT, Int64}()
for i in 1:10
dic[MyT("x", 1)] = 1
end
println(dic) # Dict(MyT("x", 1)=>1)
println(keys(dic)) # MyT[MyT("x", 1)]
Mutable values hash by identity in Julia, since without additional knowledge about what a type represents, one cannot know if two values with the same structure mean the same thing or not. Hashing mutable objects by value can be especially problematic if you mutate a value after using it as a dictionary key – this is not a problem when hashing by identity since the identity of a mutable object remains the same even when it is modified. On the other hand, it's perfectly safe to hash immutable objects by value – since they cannot be mutated, and accordingly that is the default behavior for immutable types. In the given example, if you make MyT immutable you will automatically get the behavior you're expecting:
immutable MyT # `struct MyT` in 0.6
A::String
B::Int64
end
dic = Dict{MyT, Int64}()
for i in 1:10
dic[MyT("x", 1)] = 1
end
julia> dic
Dict{MyT,Int64} with 1 entry:
MyT("x", 1) => 1
julia> keys(dic)
Base.KeyIterator for a Dict{MyT,Int64} with 1 entry. Keys:
MyT("x", 1)
For a type holding a String and an Int value that you want to use as a hash key, immutability is probably the right choice. In fact, immutability is the right choice more often than not, which is why the keywords introducing structural types has been change in 0.6 to struct for immutable structures and mutable struct for mutable structures – on the principle that people will reach for the shorter, simpler name first, so that should be the better default choice – i.e. immutability.
As #ntdef has written, you can change the hashing behavior of your type by overloading the Base.hash function. However, his definition is incorrect in a few respects (which is probably our fault for failing to document this more prominently and thoroughly):
The method signature of Base.hash that you want to overload is Base.hash(::T, ::UInt).
The Base.hash(::T, ::UInt) method must return a UInt value.
If you are overloading Base.hash, you should also overload Base.== to match.
So this would be a correct way to make your mutable type hash by value (new Julia session required to redefine MyT):
type MyT # `mutable struct MyT` in 0.6
A::String
B::Int64
end
import Base: ==, hash
==(x::MyT, y::MyT) = x.A == y.A && x.B == y.B
hash(x::MyT, h::UInt) = hash((MyT, x.A, x.B), h)
dic = Dict{MyT, Int64}()
for i in 1:10
dic[MyT("x", 1)] = 1
end
julia> dic
Dict{MyT,Int64} with 1 entry:
MyT("x", 1) => 1
julia> keys(dic)
Base.KeyIterator for a Dict{MyT,Int64} with 1 entry. Keys:
MyT("x", 1)
This is kind of annoying to do manually, but the AutoHashEquals package automates this, taking the tedium out of it. All you need to do is prefix the type definition with the #auto_hash_equals macro:
using AutoHashEquals
#auto_hash_equals type MyT # `#auto_hash_equals mutable struct MyT` in 0.6
A::String
B::Int64
end
Bottom line:
If you have a type that should have value-based equality and hashing, seriously consider making it immutable.
If your type really has to be mutable, then think hard about whether it's a good idea to use as a hash key.
If you really need to use a mutable type as a hash key with value-based equality and hashing semantics, use the AutoHashEquals package.
You did not do anything wrong. The difference between the languages is in how they choose to hash a struct when using it as a key in the map/Dict. In go, structs are hashed by their values rather than their pointer addresses. This allows programmers to more easily implement multidimensional maps by using structs rather than maps of maps. See this blog post for more info.
Reproducing Julia's Behavior in Go
To reproduce Julia's behavior in go, redefine the map to use a pointer to MyT as the key type:
func main() {
dic := make(map[MyT]int)
pdic := make(map[*MyT]int)
for i := 1; i <= 10; i++ {
t := MyT{"x", 1}
dic[t] = 1
pdic[&t] = 1
}
fmt.Println(dic)
fmt.Println(pdic)
}
Here, pdic uses the pointer to a MyT struct as its key type. Because each MyT created in the loop has a different memory address, the key will be different. This produces the output:
map[{x 1}:1]
map[0x1040a140:1 0x1040a150:1 0x1040a160:1 0x1040a180:1 0x1040a1b0:1 0x1040a1c0:1 0x1040a130:1 0x1040a170:1 0x1040a190:1 0x1040a1a0:1]
You can play with this on play.golang.org. Unlike in Julia (see below), the way the map type is implemented go means you cannot specify a custom hashing function for a user-defined struct.
Reproducing Go's Behavior in Julia
Julia uses the function Base.hash(::K, ::UInt) to hash keys for its Dict{K,V} type. While it doesn't explicitly say so in the documentation, the default hashing algorithm uses the output from object_id, as you can see in the source code. To reproduce go's behavior in Julia, define a new hash function for your type that hashes the values of the struct:
Base.hash(t::MyT, h::Uint) = Base.hash((t.A, t.B), h)
Note that you should also define the == operator in the same way to guarantee hash(x)==hash(y) implies isequal(x,y), as mentioned in the documentation.
However, the easiest way to get Julia to act like go in your example is to redefine MyT as immutable. As an immutable type, Julia will hash MyT by its value rather than its object_id. As an example:
immutable MyT
A::String
B::Int64
end
dic = Dict{MyT, Int64}()
for i in 1:10
dic[MyT("x", 1)] = 1
end
dic[MyT("y", 2)] = 2
println(dic) # prints "Dict(MyT("y",2)=>2,MyT("x",1)=>1)"
Edit: Please refer to #StefanKarpinski's answer. The Base.hash function must return a UInt for it to be a valid hash, so my example won't work. Also there's some funkiness regarding user defined types which involves recursion.
The reason you get 10 different keys is due to the fact that Julia uses the hash function when determining the key to a dict. In this case, I'm guessing that it's using the address of the object in memory as the key for the dictionary. If you'd like to explicitly make (A,B) the unique key, you'll need to override the hash function for your particular type, with something like this:
Base.hash(x::MyT) = (x.A, x.B)
That will replicate the Go behavior, with only one item in the Dict.
Here's the documentation to the hash function.
Hope that helps!