perm filename TUTOR.DOC[DOC,AIL] blob
sn#071737 filedate 1973-11-10 generic text, type T, neo UTF8
LEAP TUTORIAL
By Jim Low
LEAP TUTORIAL November 10, 1973
I. INTRODUCTION
SAIL contains an associative data system called LEAP. It is patterned
after the LEAP system implemented at LINCOLN LABORATORY by ROVNER and
FELDMAN. Features contained in our LEAP but not in the Lincoln LEAP
include a data-type called list, and "matching" procedures to be
described below.
This document is intended to serve as a companion volume to the SAIL
manual and hopefully will be easier to understand than the manual as
here we can afford expound more on the various constructs and we also
have the space to include more examples.
Other documents which may be of interest to the LEAP user include
LEAP.WRU[DOC,AIL], which is a general guide to the leap runtime
environment; LEAP.TXT[DOC,AIL], which is a detailed guide to the LEAP
parts of the SAIL compiler and the SAIL runtime system; and of course
the SAIL manual.
1
LEAP TUTORIAL November 10, 1973
II. ITEMS
The basic entities which LEAP manipulates are called "items". An item
is similar in many respects to a LISP atom. It optionally has a
printname and a datum. A datum is a scalar or an array of any SAIL
data-type other than types "item" and "itemvar". An itemvar is simply
a variable whose values are items.
As an example of an item we may consider the following declaration.
REAL ITEM RITEM;
This declares an item named RITEM whose datum is a scalar real
variable.
In addition to declarations of items at compile-time, we may
dynamically create new items by calling the function NEW. This
function may either have no arguments, (in which case the created
item has no datum); or it may have a single argument which is either
an expression or an array. This argument is copied and the copy
becomes the value of the datum of the new item. We may of course
later change the value of the datum or an element of the datum (if
the datum is an array) by using standard algolic assignments. The
data-type of the datum of the new item is the data-type of the
argument to NEW. Thus NEW(1) would create a new integer item whose
datum was initially given the value 1.
Items may be assigned to itemvars by standard SAIL assignment
statements and assignment expressions.
itmvr← itmexpr;
Items themselves are considered to be constants and thus may not
appear on the left hand side of an assignment statement or
expression.
2
LEAP TUTORIAL November 10, 1973
III. DATUMS
Associated with most items are datums which may be treated as
standard SAIL variables. To refer to the datum of an item we use the
operator DATUM.
Example:
INTEGER ITEM II;
INTEGER I;
L1: DATUM(II)←5;
L2: I ← DATUM(II)+1;
At L1 the datum of the item II is given the value "5". At L2, the
value of the datum of II is used in an arithmetic assignment
statement which would cause the variable I to receive the value 6.
Datum takes as its argument a typed item expression: Typed item
expressions include:
1. Typed items and itemvars (declared with their type followed
by ITEM or ITEMVAR) as in:
INTEGER ITEM JJ;
INTEGER ARRAY ITEM X[1:5];
INTEGER ITEMVAR ARRAY Y[1:5];
INTEGER ARRAY ITEMVAR ARRAY Z[1:5];
INTEGER ARRAY ITEMVAR Q;
Note the distinctions between the later four declarations.
X is declared to be a single item whose datum is an integer
array containing five elements. Y is declared to be an array
of five itemvars, each of which is claimed to contain an
item whose datum is a scalar integer. Z is declared to be an
array of five itemvars whose values are claimed to be items
with datums which are integer arrays. Q is an itemvar which
supposedly contains an item whose datum in an integer array.
As is shown above, we do not specify the dimensions of the
the array which is the datum of an array itemvar. Thus for
example, each element of Z could contain items whose datums
were arrays of different dimensions. With declared array
items we must declare the dimension , otherwise the
compiler would not know how much space to allocate for the
array. We place the dimensions of the array following the
item name. This is somewhat confusing as it appear that we
3
LEAP TUTORIAL November 10, 1973
have an array of items rather than a single item whose datum
is an array. SAIL has solved this problem in a very
arbitrary way by outlawing declarations of arrays of items.
One can get the effect of arrays of items by declaring
itemvar arrays and then assigning "NEW" items to the
individual elements.
2. Typed itemvar function calls:
STRING ITEMVAR PROCEDURE SNEW(STRING X);
RETURN(NEW(X));
Thus we may talk of DATUM(SNEW("anystring"))
3. Assignment expressions whose left hand side is a typed
itemvar.
The type of the datum variable is the type of its item expression.
Thus, the datum of an integer item expression is treated as an
integer variable, the datum of a real array item expression is
treated as a real array and so forth.
NOTE: no check is actually made that the item is of the claimed type.
Thus , for example, disastrous things may happen if one uses DATUM
on a string itemvar in which an integer item has been stored. The
user should be careful about storing typed items into different type
itemvars. When in doubt about the actual type an item expression he
should use the function TYPEIT to verify that the item is of the
required type. TYPEIT is a predeclared integer function.
INTEGER PROCEDURE TYPEIT(ITEMVAR X);
4
LEAP TUTORIAL November 10, 1973
The values returned by TYPEIT are:
0 - deleted or never allocated 1 - item has no datum.
2 - bracketed triple(no datum) 3 - string item
4 - real item 5 - integer item
6 - set item 7 - list item
8 - procedure item 9 - process item
10 - event-type item 11 - context item
12 - refitm item 16 - string array item
17 - real array item 18 - integer array item
19 - set array item 20 - list array item
24 - context array item 25 - invalid type(error in LEAP)
Those codes not mentioned (13-15,21-23) are also invalid and should
be considered erroneous.
5
LEAP TUTORIAL November 10, 1973
IV. SETS
A set is an unordered collection of unique items. All set variables
are initialized to PHI, the empty set consisting of no items. Set
variables receive values by assignment or by placing individual items
in them by PUT statements.
SET EXPRESSIONS:
1. Explicit Set - A sequence of item expressions which make up
the set surrounded by set brackets.
E. G.
a) {item1,item2,item3}
b) {item2,item1,item3}
c) {item2,item2,item2,item3}
Note: Since sets are unordered and a given item may appear at
most once within a set, set expressions a,b,and c above all
represent the same set.
2. PHI - the empty set. The set consisting of no elements at
all is the empty set which may be written as either
{} or PHI
3. Set Union - written SET1 ∪ SET2.
The resultant set contains all items which are elements of
either SET1 or SET2 or both.
E.G.
{item1,item2} ∪ {item2,item3} = {item1,item2,item3}
4. Set Intersection - written SET1 ∩ SET2
The resultant set contains all items which are elements of
both SET1 and SET2.
E. G.
{item1,item2,item3} ∩ {item1,item2,item4} = {item1,item2}
6
LEAP TUTORIAL November 10, 1973
5. Set Subtraction - written SET1 - SET2
The resultant set contains all items which are elements of
SET1 but not elements of SET2.
E. G.
{item1,item2,item3} - {item2,item4,item5} = {item1,item3}
PUT and REMOVE
1. To place a single item into a set variable we may use the PUT
statement:
PUT itemexpr IN setvar;
This has the identical effect as:
setvar ← setvar ∪ {itemexpr};
However, as the assignment causes the set to be copied, and
the PUT doesn,'t the PUT statement will take less time and
space to execute.
2. To remove a single item from a set variable we may use the
REMOVE statement
REMOVE itemexpr FROM setvar;
This has the same effect as:
setvar ← setvar - {itemexpr};
Again, as the REMOVE statement avoids copying the set, it is
more efficient than the equivalent assignment statement.
7
LEAP TUTORIAL November 10, 1973
SET BOOLEANS
1. Set membership
itemexpr ε setexpr
TRUE only if the item is an element of the set.
2. Set equality
setexpr1 = setexpr2
TRUE only if each item in setexpr1 is in setexpr2 and vice
versa.
3. Set inequality
setexpr1 ≠ setexpr2
TRUE if setexpr1 or setexpr2 contains an item not found in
the other.
Equivalent to
¬(setexpr1=setexpr2)
4. Proper containment
setexpr1 < setexpr2 or setexpr2 > setexpr1
TRUE if every item in setexpr1 is also in setexpr2, but
setexpr1 ≠ setexpr2 Equivalent to: ((setexpr1 ∩ setexpr2) =
setexpr1) ∧ (setexpr1 ≠ setexpr2)
5. Containment
setexpr1 ≤ setexpr2 or setexpr2 ≥ setexpr1
TRUE if every item in setexpr1 in also in setexpr2.
Equivalent to
(setexpr1 = setexpr2) ∨ (setexpr1 < setexpr2)
8
LEAP TUTORIAL November 10, 1973
COP, LOP and LENGTH
1. COP (setexpr) - returns an item which is an element of the
set. As sets are unordered you do not know which element of
the set will be returned It is useful most often when we know
the set has but a single element in which case it will return
that item.
2. LOP (setvar) - same as COP except argument must be a set
variable and the item returned is also removed from that set.
It is logically equivalent to the following procedure:
ITEMVAR PROCEDURE LOP(REFERENCE SET Y);
BEGIN ITEMVAR Q;
Q ← COP(Y);
REMOVE Q FROM Y;
RETURN(Q)
END;
LOP is often valuable if we wish some operation to be
performed on each item of a set. Consider the example below
where we wish the datums of all integer items contained in a
set SETI to be incremented by one. Assume that we have
declared IITMVR to be an integer itemvar and TSET to be a set
variable which we will use as temporaries.
TSET ← SETI; "Copy set of interest into temporary"
WHILE (TSET ≠ PHI) DO "loop while TSET has elements"
BEGIN IITMVR ← LOP(TSET); "remove an element from TSET"
IF TYPEIT(IITMVR) = 5 THEN "check if really integer"
DATUM(IITMVR) ← DATUM(IITMVR) + 1;
END;
NOTE: LOP is compiled into code other than a straightforward
procedure call and thus like many other functionals cannot
appear as a statement but only as part of an expression. Thus
if we just wanted to remove an arbitrary set element and
throw if away we would have to say:
DMY ← LOP(SETVAR);
where DMY is an itemvar whose contents we do not care about,
rather than the simpler:
LOP(SETVAR);
9
LEAP TUTORIAL November 10, 1973
3. LENGTH(setexpr) - returns the number of items within a set.
Logically equivalent to following procedure:
INTEGER PROCEDURE LENGTH(SET Y);
BEGIN "LENGTH"
INTEGER COUNT; ITEMVAR DUMMY;
COUNT ← 0;
WHILE (Y ≠ PHI) DO
BEGIN
DUMMY ← LOP(Y);"remove an element from the set"
COUNT ← COUNT +1; "step count of elements"
END;
RETURN(COUNT);
END "LENGTH";
The actual implementation of LENGTH is much more efficient
than the above procedure (usually taking only two machine
instructions). The most efficient way of determining if a
given set is empty is to see if the LENGTH of that set is
zero. This is much faster that comparing the set and PHI for
equality.
10
LEAP TUTORIAL November 10, 1973
V. LISTS
A list is an ordered sequence of items (not necessarily distinct).
List variables are initialized to NIL the empty list containing no
items. List variables receive values by assignment or by placing
individual items in them by PUT statements.
LIST Expressions
1. Explicit List - written as the sequence of items (separated
by commas) all surrounded by list brackets "{{ }}".
a. {{ item1, item2, item3}}
b. {{ item2, item1, item3}}
c. {{ item2, item2, item1, item 3}}
Note that since the order and number of times each item
appears is important with lists, the expressions a, b, and c
above all represent different list expressions.
NOTE: we may represent NIL, the empty list, by {{ }}
2. Concatenation - written list1 & list2. This forms a new
list containing all the items in list1 followed by all the
items in list2.
E. G.
{{item1, item2, item3,}} & {{item3, item4, item5 }}
=
{{item1, item2, item3, item3, item4, item5}}
{{item1, item2}} & {{item 4, item4, item5}}
=
{{item1, item2, item4, item4, item5}}
11
LEAP TUTORIAL November 10, 1973
3. Sublists - There are two forms of sublist expressions
a. listexpr [i1 TO i2] - the first integer expression (i1)
stands for the position of the first element to be taken
and the second (i2) stands for the position of the last
element to be taken.
E. G.
{{itema,itemb,itemc,itemd}}[2 TO 3]={{itemb,itemc}}
{{itema,itemb,itemc,itemd}}[3 TO 3]={{itemc}}
b. listexpr [i1 FOR i2] - the first integer expression (i1)
stands for the position of the first element to be taken
and the second (i2) stands for the number of elements to
be taken. E. G.
{{itema,itemb,itemc,itemd}}[2 FOR 3]={{itemb,itemc,itemd}}
{{itema,itemb,itemc,itemd}}[3 FOR 1]={{itemc}}
{{itema,itemb,itemc,itemd}}[3 FOR 0]={{ }}= NIL
LIST Selectors
It is often useful to think of a list as an untyped itemvar array
with a single dimension with lower bound 1 and upper bound variable.
1. Expression selector
listexpr [i1] - returns the item which is the i1 element of
the list. If i1 is less than 0 or greater than the number of
elements of the list an error is signaled.
E. G.
{{itema, itemb, itemc}} [1] = itema
{{itemb, itemc, itemd}} [2] = itemc
Note the difference between listexpr[i1] and listexpr[i1 FOR
1]. The former returns an item and the later returns a list
containing a single item.
12
LEAP TUTORIAL November 10, 1973
2. Replacement selector
listvar[i1] ← itemexpr;
This removes the i1 element of the list and replaces it with
the itemexpr i1 must be between 1 and the number of elements
in the list + 1.
E. G.
LIST1 ← {{ITEM1}};
LIST1[1] ← ITEM2; "NOW LIST1 = {{ITEM2}}"
LIST1[2] ← ITEM3; "NOW LIST1 = {{ITEM2,ITEM3}}"
LIST1[1] ← LIST1[2]; "NOW LIST1 = {{ITEM3,ITEM3}}"
13
LEAP TUTORIAL November 10, 1973
LIST BOOLEANS
itm ε listexpr - TRUE if the itm is an element of the listexpr
listexpr1 = listexpr2 - TRUE if the lists contain the same
items in the same order. LIST FUNCTIONS
LENGTH(listexpr) - this returns the current length of the list.
COP(listexpr) - this is identical in semantics with the
expression "listexpr[1]"
LOP(listvar) - this returns the first item in the list
variable and as a side effect removes the first element from
the list variable. It is logically equivalent to the
following procedure.
ITEMVAR PROCEDURE LOP(REFERENCE LIST ARG); BEGIN
"LOP" ITEMVAR RESULT; RESULT ←
ARG[1]; ARG ← ARG[2 TO LENGTH(ARG)];
RETURN(RESULT); END "LOP";
LISTX(listexpr,itemexpr,intexpr) - this integer function
returns 0 if the "itemexpr" does not appear in the "listexpr"
at least "intexpr" different times. Otherwise it returns the
index of the "intexpr" occurence of the "itemexpr" within the
"listexpr". Thus,
LISTX({{FOO,BAZ,GARP,BAZ,BAZ}},BAZ,2) is 4.
LISTX({{FOO,BAZ,GARP,BAZ,BAZ}},GARP,2) IS 0.
14
LEAP TUTORIAL November 10, 1973
LIST PUT AND REMOVE STATEMENTS
When we insert items into a list we need to specify in some way the
position within the list where the item is to be placed. We do this
by specifying either an index position, or item BEFORE or AFTER
which the item is to be inserted.
PUT itm1 IN listvar AFTER n - will insert "itm1" after the "nth"
element of the "listvar". Thus to insert an item into the
list after every element in the list we would execute:
PUT itm1 IN listvar AFTER LENGTH(listvar);
To insert the item at the head of the list we would use the index 0:
PUT itm1 IN listvar AFTER 0;
If the index is less than 0 or greater than the length of the list
variable then an error message is generated.
PUT itm IN listvar BEFORE n - this is identical in semantics
with:
PUT itm IN listvar AFTER N-1;
PUT itm1 IN listvar AFTER itm2;-- this statement searches
for the first occurrence of itm2 within the listvar and
inserts itm1 into the list following itm2. If itm2 is not an
element of the list then itm1 is inserted at the end of the
list. It is equivalent to :
PUT itm1 IN listvar AFTER (IF LISTX(listvar,itm2,1)
≠ 0 THEN LISTX(listvar,itm2,1) ELSE LENGTH(listvar);
PUT itm1 IN listvar BEFORE itm2 - this statement searches the
listvar for the first occurence of itm2 and then inserts itm1
ahead of itm2 in the list. It is logically equivalent to:
PUT itm1 IN listvar BEFORE (IF LISTX(listvar,itm2,1)≠ 0
THEN LISTX(listvar,itm2,1) ELSE 1;
15
LEAP TUTORIAL November 10, 1973
REMOVE intexpr FROM listvar -- This removes the "intexpr" element
from the listvar. It is equivalent to:
listvar ← listvar[0 to intexpr-1] &
listvar[intexpr+1 to LENGTH(listvar)];
REMOVE itm FROM listvar - This removes the first occurrence
of "itm" from the list variable. It is equivalent to:
IF itm ε listvar THEN REMOVE LISTX(listvar,itm,1) FROM listvar;
REMOVE ALL itm FROM listvar - This removes all instances of
"itm" from tthe listvar. It is equivalent to:
WHILE itm ε listvar DO REMOVE itm FROM listvar;
16
LEAP TUTORIAL November 10, 1973
VI. ASSOCIATIONS
The associative power of LEAP comes from the use of associations,
also called triples or associative triples. A triple is a 3-tuple of
items. We write a triple as:
A⊗O≡V
where A, O, and V are items or item expressions. We call the first
component of the triple (A above) the "attribute"; the second
component (O above), the "object"; and the third component (V above),
the "value". Triples are kept in the "associative store". Triples are
inserted into the associative store by MAKE statements and removed
from the associative store by ERASE statements.
MAKE
A MAKE statement is of the form:
MAKE itmexpr1⊗itmexpr2 ≡itemexpr3;
If the triple already exists in the associative store, the statement
does nothing, otherwise the triple is inserted into the store.
ERASE
To remove a triple from the associative store we execute an ERASE
statement:
ERASE itmexpr1⊗itmexpr2≡ itemexpr3;
If the triple is not in the associative store, the statement does
nothing, otherwise the triple is removed from the associative store.
We often wish to erase all the triples which have specific items as 1
or 2 of their components but we don't care about the remaining
components. To do this we may use the token ANY to stand for the
unspecified components.
17
LEAP TUTORIAL November 10, 1973
E. G. to erase all associations with item2 as their object we could
use:
ERASE ANY⊗item2≡ANY;
to erase all associations with item1 as their attribute and item2 as
their object we would use:
ERASE item1⊗item2 ≡ ANY;
ANY may be used in 0, 1, 2, or all 3 positions in the triple. Thus,
ERASE ANY⊗ANY≡ANY;
would get rid of all associations in the UNIVERSE.
18
LEAP TUTORIAL November 10, 1973
ASSOCIATIVE BOOLEANS
We may determine if a given triple exists by using the boolean
expression:
itmexp1 ⊗ itmexp2 ≡ itmexp3
which will evaluate to TRUE if the triple containing the items exists
in the associative store. As with ERASE 1 or 2 of the components may
be ANY. Thus,
ANY ⊗ ANY ≡ item1
will yield the value TRUE if any triple in the associative store
contains item1 as its value component.
BINDING ASSOCIATIVE BOOLEANS
Another form of the associative boolean takes one or more itemvars
(prefaced by the operator BIND) in place of corresponding item
expressions. The boolean value of this expression is the same as
that of the associative boolean formed by replacing the "BIND
ITEMVAR" terms with ANY. thus the expression:
FATHER ⊗ JOHN ≡ BIND itv
would have the same truth value as the boolean expression:
FATHER ⊗ JOHN ≡ ANY
The binding boolean though, has an additional side effect if the
value returned is TRUE. In that case it assigns to the itemvar
(prefaced by BIND) the value of an item which makes the associative
boolean TRUE. For example above, the interpreter would search to see
if there was any triple in the associative store, which had both
FATHER as its attribute component and JOHN as its object component.
If it found one, the interpreter would then assign the item in the
value component to the itemvar "itv".
Note that if there is more than one triple which satisfies the
associative boolean, which one is selected is undefined. If there is
no triple which satisfies the boolean then the boolean returns FALSE
19
LEAP TUTORIAL November 10, 1973
and the value of the BIND itemvar remains unchanged. Thus if we had a
group of associations: with attributes, the items CAR and CDR
(representing a LISP like structure) we could find the last cell in
the CDR direction of the LISP-list pointed to by first, using the
very simple program:
LAST ← FIRST ;
WHILE (CDR ⊗ LAST ≡ BIND LAST ) DO;
Note that each time the associative boolean is executed, the current
value of the itemvar LAST is used as the object component, and if
successful the value of LAST is changed to become the next element in
the CDR direction.
20
LEAP TUTORIAL November 10, 1973
DERIVED SETS
In order to use the associative nature of triples we must have ways
of finding triples which have certain specified components. One way
we have discussed above is the use of binding associative booleans.
Another is to use derived sets. The third, FOREACH statements, will
be discussed later.
There are three forms of derived sets now implemented: the (')
derived set, the (⊗) derived set, and the (≡) derived set.
itmexp1 ' itmexp2
produces the set of all items X, such that
itmexp1 ⊗ X ≡ itmexp2
is a triple currently in the associative store.
itmexp1 ⊗ itmexp2
produces the set of all items Y such that,
itmexp1 ⊗ itmexp2 ≡ Y
is a triple in the associative store.
itmexp1 ≡ itmexp2
produces the set of all items Y such that.
Y ⊗ itmexp1 ≡ itmexp2
is a triple in the associative store
One or both of the item expressions may be the token ANY. Again this
means that we do not care about the value of that component. Thus,
itmexp1 ⊗ ANY
will search the associative store for all associations which have
itmexp1 as their attribute component, and will return the set of
value components of such associations.
21
LEAP TUTORIAL November 10, 1973
EXAMPLE -DERIVED SETS
Let us represent a family tree using associations. We will have the
declared item PARENT and the sets MALE and FEMALE, as well as items
representing members of the family: JOE, TIM, TED, JOYCE, JANET,
ALICE, and HARRIET.
The familial relationships are represented by a the following tree.
TOM ALICE JOE JAN
| | | |
--------------- ------------
| |
------------- |
| | |
JOYCE TED TIM
| |
-------------------------------------
|
HARRIET
Thus, the parents of HARRIET are JOYCE and TIM; the parents of JOYCE
and TED are TOM and ALICE and so forth.
We can represent this tree by making the following associations:
MAKE PARENT ⊗ HARRIET ≡ JOYCE;
MAKE PARENT ⊗ HARRIET ≡ TIM;
MAKE PARENT ⊗ JOYCE ≡ TOM;
MAKE PARENT ⊗ JOYCE ≡ ALICE;
MAKE PARENT ⊗ TED ≡ ALICE;
MAKE PARENT ⊗ TED ≡ TOM;
MAKE PARENT ⊗ TIM ≡ JOE;
MAKE PARENT ⊗ TIM ≡ JAN;
To keep track of the sexes of the various people we have the two sets
MALE and FEMALE.
MALE ← {TIM,TED,TOM,JOE};
FEMALE ← {JAN,JOYCE,ALICE};
NOTE: The above is merely one possible way we might represent the
family tree. For example instead of the MALE and FEMALE sets, we
22
LEAP TUTORIAL November 10, 1973
might have associations of the form: SEX ⊗ person ≡ MALE, where MALE
is now an item. One of the interesting difficulties in using LEAP is
deciding how to represent a given system of data as LEAP will often
allow many different ways of representing the same information. Some
of the tradeoffs between the different representations will be
discussed later.
Now to use the structure we have built. Let us say that we wished to
find the parents of Harriet. We may easily do this by use of a
derived set.
HARRIET_PARENTS ← PARENTS ⊗ HARRIET;
where HARRIET_PARENTS has been declared to be a set variable.
To find Harriet's brothers is a little more complicated.
"find one parent"
PARENT_ITMVR ← COP(PARENTS ⊗ HARRIET);
Comment the above could be replaced by
the binding associative boolean.
IF ¬(PARENTS⊗HARRIET≡BIND PARENT_ITMVR) THEN
PRINT("HARRIET IS AN ANDROID")
;
"set of brothers,sisters"
SIBLING_SET ← PARENT ' PARENT_ITMVR;
"brother is a male sibling"
BROTHER_SET ← SIBLING_SET ∩ MALES;
The above example illustrates the use of associations as binary
relations between items, in this case the relation "parent of". Often
we wish to associate several different pieces of data with an item.
To do this we may declare items which will be used to name the data
and then allocate items which will contain the corresponding data for
each item. For example we may wish to record such various attributes
of a person such as weight, height, nickname. To do this we will have
items WEIGHT, HEIGHT, and NICKNAME which will be used to name the
attributes. We will allocate items whose datums are the corresponding
values. E.G.
MAKE WEIGHT ⊗ JOE ≡ NEW(165);
MAKE HEIGHT ⊗ JOE ≡ NEW (70);
MAKE NICKNAME ⊗ JOE ≡ NEW("JOEY");
Then to find the value of an attribute such as weight we would use
the expression:
DATUM(INT_ITMVR ← COP(WEIGHT⊗JOE))
23
LEAP TUTORIAL November 10, 1973
Remember that the assignment of the item to the integer itmvr is
required so that the compiler can tell what the data type of the
datum is. Again we could use a binding associative boolean instead
of a COP of a derived set. The binding boolean generates more
efficient code, but has the disadvantage of returning a boolean value
rather than an item value as does COP. However since COP will stop
your program if you apply it to an empty set, it is well worth the
syntactic bother to use the associative boolean.
FOREACH STATEMENTS
By using these operations and set variables we have sufficient power
to do any search on the associative data base. However one soon
realizes that they are very inconvenient to use in all but the most
simple cases. Therefore another technique is provided called FOREACH
statements.
A FOREACH statement is similar to a FOR statement in that it causes
the iteration of a given SAIL statement to be performed with a
control variable receiving various values for each iteration. These
values are obtained by searching the associative data base or
enumerating the members of a set of items.
The most simple FOREACH statement has a single "local" itemvar and a
single "associative context". A local itemvar serves the same purpose
as the loop variable in a FOR statement. With each iteration it will
receive an item value and a SAIL statement will be executed. A simple
example of a FOR statement is:
FOREACH X | BROTHER⊗BOY1≡ X DO
<stmt>
This statement is equivalent to the following:
listx← BROTHER⊗BOY;
FOR j ← 1 step 1 until LENGTH(LISTX) DO
BEGIN X←listx[j];
<stmt>
END;
ANY and BINDIT
There are two special "items" which may not be arguments to a make
statement (and thus may not be components of any association in the
associative data base). These are ANY and BINDIT. These items may
appear anywhere else an item might be legal, such as within sets,
itemvars and lists.
24
LEAP TUTORIAL November 10, 1973
We have mentioned ANY earlier. It essentially represents "don't care"
when we are doing pattern matching on the associative store. Thus
FOREACH x | x ⊗ANY ≡ B DO
is to be iterated once for every distinct X which appears in
associations with B as the value component.
BINDIT is used to conditionally control ? searches. It also has
importance in MATCHING PROCEDURES which will be discussed later.
There are special forms of the BINDING BOOLEANS and FOREACH
statements, are affected when an itemvar to be bound contains BINDIT.
The BINDING BOOLEAN B ⊗ C ≡ ? x is identical in effect with the
expression
(IF x = BINDIT THEN (B⊗C≡ BIND x) ELSE (B⊗C≡x))
That is the value of x is used if not equal to BINDIT and the
itemvar x is to be bound if the value of x is equal to BINDIT.
FOREACH ?x,y | A⊗x≡ y DO <stmt>
is similarly equivalent to
IF X=BINDIT THEN
FOREACH x,y | A⊗x=y DO <stmt>
else FOREACH y | A⊗x≡y DO <stmt>
Note that the FOREACH in the else part of the statement does not have
"x" in its binding list and thus x is considered to have a value. The
reverse is true in the THEN part of the statement.
25