Tuples are one dimension arrays. (,) is an empty tuple, (1,) is one item tuple, and (1,2) is two item tuple. We can include a tuple in a tuple: ((1,2), (3,4)) is a tuple with two tuples. We can use a pointer to array to hold a tuple. We can change values using Return statement, using 0 for first item always. Using ++ -- += -= *= /* -! which affect all items (not those who have objects or strings). To access an item we can use Array(Pointer2Array, 2) to get 3rd item.
We can make array pointers for multidimensional arrays, and with different bases, one for all or different value for each dimension. So Array() works using the proper base(s) and dimensions.
We see in the example below, that we can make new references A1 to C1 for A to C tuples. We can make copies when we make empty arrays A(),B$(), C() and assign values as A, B, C. We see that C() is a shallow copy of C, which means is a new array, but any pointer in the element is a copy of pointer so we get the same objects. We can change any of this, just assign a new pointer, or we can change the array which points the pointer, so we get new values for each copy.
Second part show how we use pointers to multi dimension arrays. Also we see how we change dimension preserving items and how to add items.
Using objects (Groups) is not the same as tuples. We can use Groups without pointers, or to be precise with a single pointer, which means that we get not a copy of a pointer but always a new pointer, a deep copy of group. See example TestGroup, A(0).K() is different from B(0).K()
For advanced readers:
We can get a pointer to a Group (internal as a weak reference) or to a copy of group (internal as true pointer), and copy it to an array element. So in TestGroup2 we have pointer to group, and for that reason we get a shallow copy, so when we change the group in A() we get the change in B() two.
Using ->(A) we get pointer from a copy of A, and using ->A we get reference to A. The second is valid until Group A erased (at the exit of module TestGroup2). In Functions we have = as identifier to return value (Function Square(x) {=x**2}) but we can use -> to return pointer (we have to return a true pointer, so something like ->A is a fault, but not for the time we return the pointer, but when we use it, and interpreter can't resolve to actual item).
\\ Pointers for multi dimension arrays
Dim A(2 to 10, 5 to 10)=1
M=A()
Print Len(M) = 54 ' 9X6
Print array(M, 2,5)=1 \\ M point to a 2 dimension Array
\\ Return use one dimension, so 0 is the first element
Return M, 0:=1000, 6:=5000 ' 6 is the 7th item, first in second row (row, columns)
Print array(M, 2,5)=1000, array(M, 3,5)
Dim A(2 to 11, 5 to 10) ' add one row
Print Len(M) = 60 ' 10X6
Print Type$(A(11, 5))="Empty" ' new raw has Empty as value (in calculations this is same as 0 or empty string)
\\ So now we put a value
For i=5 to 10:A(11,i)=1:Next i
\\ we can alter last item using pointer M
Return M, 59:=9999
Print A(11, 10)=9999
\\ assuming we have defalut base 0
Dim A(10,6) ' redim preserving values
Print A(9, 5)=9999, Len(A())=60
\\ we can use Base 1 or Base 0 to explicit declare base
Dim Base 1, A(10,6) ' redim preserving values
Print A(10, 6)=9999, Len(A())=60
\\ or we can use for each dimension a new base
Dim A(5 to 14, 10 to 15) ' redim preserving values
Print A(14, 15)=9999, Len(A())=60
\\ Get dimensions, width for each dimension, base (min value) for each dimension, max value for each dimension
Print Dimension(A())=2 ' 2 dimension
Print Dimension(A(),0)=5 ' first dimension base is 5
Print Dimension(A(),1)=10 ' 10 items
Print Dimension(A(),2)=6 ' 6 items, so we have 10x6 items
Print Dimension(A(),1,0)=5
Print Dimension(A(),1,1)=14
Print Dimension(A(),2,0)=10
Print Dimension(A(),2,1)=15
\\ Copy all item to stack using pointer to array (not A() but M)
Flush ' now stack is empty stack
Push ! M ' now get 60 items
Print stack.size=60 ' true
Stack ' now display all stack items
Flush ' now empty stack
\\ if we use Push ! M we send values in reverse
Data ! M ' now get 60 items
\\ now all item make an array and return a pointer to Z
Z=Array([])
Print stack.size=0
Link Z to Z()
Dim Z(5 to 14, 10 to 15)
Print Z()
Print Z(14,15)=9999
Z++ ' Add 1 to all items
Print Z()
Z(14,15)-- ' Subtract 1 from one item
Print Z(14,15)
Module TestGroup {
Group A {
X=10
Dim K(10)=1
}
Dim A(), B()
A()=(A,)
B()=A()
A(0).X++
A(0).K(0)=1000
Print A(0).X=11, A(0).K()
Print B(0).X=10, B(0).K()
}
TestGroup
We can make array pointers for multidimensional arrays, and with different bases, one for all or different value for each dimension. So Array() works using the proper base(s) and dimensions.
We see in the example below, that we can make new references A1 to C1 for A to C tuples. We can make copies when we make empty arrays A(),B$(), C() and assign values as A, B, C. We see that C() is a shallow copy of C, which means is a new array, but any pointer in the element is a copy of pointer so we get the same objects. We can change any of this, just assign a new pointer, or we can change the array which points the pointer, so we get new values for each copy.
Second part show how we use pointers to multi dimension arrays. Also we see how we change dimension preserving items and how to add items.
Using objects (Groups) is not the same as tuples. We can use Groups without pointers, or to be precise with a single pointer, which means that we get not a copy of a pointer but always a new pointer, a deep copy of group. See example TestGroup, A(0).K() is different from B(0).K()
For advanced readers:
We can get a pointer to a Group (internal as a weak reference) or to a copy of group (internal as true pointer), and copy it to an array element. So in TestGroup2 we have pointer to group, and for that reason we get a shallow copy, so when we change the group in A() we get the change in B() two.
Using ->(A) we get pointer from a copy of A, and using ->A we get reference to A. The second is valid until Group A erased (at the exit of module TestGroup2). In Functions we have = as identifier to return value (Function Square(x) {=x**2}) but we can use -> to return pointer (we have to return a true pointer, so something like ->A is a fault, but not for the time we return the pointer, but when we use it, and interpreter can't resolve to actual item).
Form
80, 60
\\ Tuple of items (as one dimension arrays)
A=(1,2,3,4,5)
B=("George", 10, "Bob", 5)
C=(("George", 10),("Bob", 5))
Print Len(A)=5 ' true
Print Len(B)=4 ' true
Print Len(C)=2 ' true
\\ get reference of A to A1
A1=A
B1=B
C1=C
\\ Get Shallow Copy
Dim A(), B$(), C()
A()=A
B$()=B
C()=C
A(0)=10
Print A ' 1 2 3 4 5
Print A() ' 10 2 3 4 5
Link B$() to B()
B$(0)="Hello George"
B(1)=1000
Print B ' George 10 Bob 5
Print B$() ' Hello George 1000 Bob 5
C()=C
C(0)=("New Name", 500)
Print Array(C,0), Array(C, 1) ' George 10 Bob 5
Print C(0)(), C(1)() ' New Name 500 Bob 5
\\ Test Shallow Copy
\\ we keep pointer to second array but we change values
\\ we need a pointer to C[1]
N=Array(C,1)
\\ So we can use Return to return multiple values
Return N, 0:="New Bob", 1:=5000
\\ So Array(C,1) show us new values
Print Array(C,0), Array(C, 1) ' George 10 New Bob 5000
\\ And because we get shallow copy (pointer only) we get C(1)() array with new values
Print C(0)(), C(1)() ' New Name 500 New Bob 5000
\\ Now C(1) get a new pointer
C(1)=("Another Name", 2000)
Print Array(C,0), Array(C, 1) ' George 10 New Bob 5000
Print C(0)(), C(1)() ' New Name 500 Another Name 2000
\\ we can get a copy of A using Cons() with one argument
NewArray=Cons(A) ' copy of A
Print NewArray
NewArray2=Cons(A, A) ' add A twice
Print NewArray2
\\ Get a copy of C() to a pointer
CopyC=Cons(C())
Print Array(CopyC, 0), Array(CopyC, 1)
Return CopyC, 1:=("Just Another Name", 3000)
Print Array(CopyC, 0), Array(CopyC, 1) ' New Name 500 Just Another Name 3000
Print C(0)(), C(1)() ' New Name 500 Another Name 2000
\\ Tuple of items (as one dimension arrays)
A=(1,2,3,4,5)
B=("George", 10, "Bob", 5)
C=(("George", 10),("Bob", 5))
Print Len(A)=5 ' true
Print Len(B)=4 ' true
Print Len(C)=2 ' true
\\ get reference of A to A1
A1=A
B1=B
C1=C
\\ Get Shallow Copy
Dim A(), B$(), C()
A()=A
B$()=B
C()=C
A(0)=10
Print A ' 1 2 3 4 5
Print A() ' 10 2 3 4 5
Link B$() to B()
B$(0)="Hello George"
B(1)=1000
Print B ' George 10 Bob 5
Print B$() ' Hello George 1000 Bob 5
C()=C
C(0)=("New Name", 500)
Print Array(C,0), Array(C, 1) ' George 10 Bob 5
Print C(0)(), C(1)() ' New Name 500 Bob 5
\\ Test Shallow Copy
\\ we keep pointer to second array but we change values
\\ we need a pointer to C[1]
N=Array(C,1)
\\ So we can use Return to return multiple values
Return N, 0:="New Bob", 1:=5000
\\ So Array(C,1) show us new values
Print Array(C,0), Array(C, 1) ' George 10 New Bob 5000
\\ And because we get shallow copy (pointer only) we get C(1)() array with new values
Print C(0)(), C(1)() ' New Name 500 New Bob 5000
\\ Now C(1) get a new pointer
C(1)=("Another Name", 2000)
Print Array(C,0), Array(C, 1) ' George 10 New Bob 5000
Print C(0)(), C(1)() ' New Name 500 Another Name 2000
\\ we can get a copy of A using Cons() with one argument
NewArray=Cons(A) ' copy of A
Print NewArray
NewArray2=Cons(A, A) ' add A twice
Print NewArray2
\\ Get a copy of C() to a pointer
CopyC=Cons(C())
Print Array(CopyC, 0), Array(CopyC, 1)
Return CopyC, 1:=("Just Another Name", 3000)
Print Array(CopyC, 0), Array(CopyC, 1) ' New Name 500 Just Another Name 3000
Print C(0)(), C(1)() ' New Name 500 Another Name 2000
\\ Pointers for multi dimension arrays
Dim A(2 to 10, 5 to 10)=1
M=A()
Print Len(M) = 54 ' 9X6
Print array(M, 2,5)=1 \\ M point to a 2 dimension Array
\\ Return use one dimension, so 0 is the first element
Return M, 0:=1000, 6:=5000 ' 6 is the 7th item, first in second row (row, columns)
Print array(M, 2,5)=1000, array(M, 3,5)
Dim A(2 to 11, 5 to 10) ' add one row
Print Len(M) = 60 ' 10X6
Print Type$(A(11, 5))="Empty" ' new raw has Empty as value (in calculations this is same as 0 or empty string)
\\ So now we put a value
For i=5 to 10:A(11,i)=1:Next i
\\ we can alter last item using pointer M
Return M, 59:=9999
Print A(11, 10)=9999
\\ assuming we have defalut base 0
Dim A(10,6) ' redim preserving values
Print A(9, 5)=9999, Len(A())=60
\\ we can use Base 1 or Base 0 to explicit declare base
Dim Base 1, A(10,6) ' redim preserving values
Print A(10, 6)=9999, Len(A())=60
\\ or we can use for each dimension a new base
Dim A(5 to 14, 10 to 15) ' redim preserving values
Print A(14, 15)=9999, Len(A())=60
\\ Get dimensions, width for each dimension, base (min value) for each dimension, max value for each dimension
Print Dimension(A())=2 ' 2 dimension
Print Dimension(A(),0)=5 ' first dimension base is 5
Print Dimension(A(),1)=10 ' 10 items
Print Dimension(A(),2)=6 ' 6 items, so we have 10x6 items
Print Dimension(A(),1,0)=5
Print Dimension(A(),1,1)=14
Print Dimension(A(),2,0)=10
Print Dimension(A(),2,1)=15
\\ Copy all item to stack using pointer to array (not A() but M)
Flush ' now stack is empty stack
Push ! M ' now get 60 items
Print stack.size=60 ' true
Stack ' now display all stack items
Flush ' now empty stack
\\ if we use Push ! M we send values in reverse
Data ! M ' now get 60 items
\\ now all item make an array and return a pointer to Z
Z=Array([])
Print stack.size=0
Link Z to Z()
Dim Z(5 to 14, 10 to 15)
Print Z()
Print Z(14,15)=9999
Z++ ' Add 1 to all items
Print Z()
Z(14,15)-- ' Subtract 1 from one item
Print Z(14,15)
Module TestGroup {
Group A {
X=10
Dim K(10)=1
}
Dim A(), B()
A()=(A,)
B()=A()
A(0).X++
A(0).K(0)=1000
Print A(0).X=11, A(0).K()
Print B(0).X=10, B(0).K()
}
TestGroup
Module TestGroup2
{
Group A {
X=10
Dim K(10)=1
}
Dim A(), B()
A()=(0,) ' one item
A(0)->(A) ' now A(0) has a pointer to a copy of A
B()=A()
A(0).X++
A(0).K(0)=1000
Print A(0).X=11, A(0).K()
Print B(0).X=11, B(0).K()
}
TestGroup2
Group A {
X=10
Dim K(10)=1
}
Dim A(), B()
A()=(0,) ' one item
A(0)->(A) ' now A(0) has a pointer to a copy of A
B()=A()
A(0).X++
A(0).K(0)=1000
Print A(0).X=11, A(0).K()
Print B(0).X=11, B(0).K()
}
TestGroup2
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