I've found the symbol :=: in some Clarion code and I can't seem to figure out exactly what it does. The code was written by a previous developer many years ago, so I can't ask him. I also have not been able to find any results for "colon equals colon" in Google.
Here is an example of the code, where bufSlcdpaDtl is a file object:
lCCRecord Like(bufSlcdpaDtl),Pre(lCCRecord)
! ...other stuff...
lCCRecord :=: bufSlcdpaDtl
I'm wondering if it's something similar to ::= in Python or possibly the assignment operator :=.
In the language reference manual on page 561 This is called the Deep assignment operator. The syntax is destination :=: source. Destination can be a label of a GROUP, RECORD, QUEUE ds or an array. The source can be the same plus a numeric, string const, variable, procedure, or expression. It will perform multiple individual component variable assignments from one ds to another. More information can be found in that document as well as the apparent home of clarion: http://www.softvelocity.com/
A great example of what the Deep Assignment operator does:
Group1 GROUP
S SHORT
L LONG
END
Group2 GROUP
L SHORT
S REAL
T LONG
END
ArrayField SHORT,DIM(1000)
CODE
Group2 :=: Group1 ! Is equivalent to:
! Group2.S = Group1.S
! Group2.L = Group1.L
! and performs all necessary data conversion
ArrayField :=: 7 ! Is equivalent to:
! LOOP I# = 1 to 1000
! ArrayField[I#] = 7
! END
Related
My input text might have a simple statement like this:
aircraft
In my language I call this a name which represents a set of instances with various properties.
It yields an instance_set of all aircraft instances in this example.
I can apply a filter in parenthesis to any instance_set:
aircraft(Altitude < ceiling)
It yields another, possibly reduced instance_set.
And since it is an instance set, I can filter it yet again:
aircraft(Altitude < ceiling)(Speed > min_speed)
I foolishly thought I could do something like this in my grammar:
instance_set = expr
expr = source / instance_set
source = name filter?
It parses my first two cases correctly, but chokes on the last one:
aircraft(Altitude < ceiling)(Speed > min_speed)
The error reported being just before the second open paren.
Why doesn't Arpeggio see that there is just a filtered instance_set which is itself a filtered instance set?
I humbly submit my appeal to the peg parsing gods and await insight...
Your first two cases both match source. Once source is matched, it's matched; that's the PEG contract. So the parser isn't going to explore the alternative.
But suppose it did. How could that help? The rule says that if an expr is not a source, then it's an instance_set. But an instance_set is just an expr. In other words, an expr is either a source or it's an expr. Clearly the alternative doesn't get us anywhere.
I'm pretty sure Arpeggio has repetitions, which is what you really want here:
source = name filter*
I've seen := used in several code samples, but never with an accompanying explanation. It's not exactly possible to google its use without knowing the proper name for it.
What does it do?
http://en.wikipedia.org/wiki/Equals_sign#In_computer_programming
In computer programming languages, the equals sign typically denotes either a boolean operator to test equality of values (e.g. as in Pascal or Eiffel), which is consistent with the symbol's usage in mathematics, or an assignment operator (e.g. as in C-like languages). Languages making the former choice often use a colon-equals (:=) or ≔ to denote their assignment operator. Languages making the latter choice often use a double equals sign (==) to denote their boolean equality operator.
Note: I found this by searching for colon equals operator
It's the assignment operator in Pascal and is often used in proofs and pseudo-code. It's the same thing as = in C-dialect languages.
Historically, computer science papers used = for equality comparisons and ← for assignments. Pascal used := to stand in for the hard-to-type left arrow. C went a different direction and instead decided on the = and == operators.
In the statically typed language Go := is initialization and assignment in one step. It is done to allow for interpreted-like creation of variables in a compiled language.
// Creates and assigns
answer := 42
// Creates and assigns
var answer = 42
Another interpretation from outside the world of programming languages comes from Wolfram Mathworld, et al:
If A and B are equal by definition (i.e., A is defined as B), then this is written symbolically as A=B, A:=B, or sometimes A≜B.
■ http://mathworld.wolfram.com/Defined.html
■ https://math.stackexchange.com/questions/182101/appropriate-notation-equiv-versus
Some language uses := to act as the assignment operator.
In a lot of CS books, it's used as the assignment operator, to differentiate from the equality operator =. In a lot of high level languages, though, assignment is = and equality is ==.
This is old (pascal) syntax for the assignment operator. It would be used like so:
a := 45;
It may be in other languages as well, probably in a similar use.
A number of programming languages, most notably Pascal and Ada, use a colon immediately followed by an equals sign (:=) as the assignment operator, to distinguish it from a single equals which is an equality test (C instead used a single equals as assignment, and a double equals as the equality test).
Reference: Colon (punctuation).
In Python:
Named Expressions (NAME := expr) was introduced in Python 3.8. It allows for the assignment of variables within an expression that is currently being evaluated. The colon equals operator := is sometimes called the walrus operator because, well, it looks like a walrus emoticon.
For example:
if any((comment := line).startswith('#') for line in lines):
print(f"First comment: {comment}")
else:
print("There are no comments")
This would be invalid if you swapped the := for =. Note the additional parentheses surrounding the named expression. Another example:
# Compute partial sums in a list comprehension
total = 0
values = [1, 2, 3, 4, 5]
partial_sums = [total := total + v for v in values]
# [1, 3, 6, 10, 15]
print(f"Total: {total}") # Total: 15
Note that the variable total is not local to the comprehension (so too is comment from the first example). The NAME in a named expression cannot be a local variable within an expression, so, for example, [i := 0 for i, j in stuff] would be invalid, because i is local to the list comprehension.
I've taken examples from the PEP 572 document - it's a good read! I for one am looking forward to using Named Expressions, once my company upgrades from Python 3.6. Hope this was helpful!
Sources: Towards Data Science Article and PEP 572.
It's like an arrow without using a less-than symbol <= so like everybody already said "assignment" operator. Bringing clarity to what is being set to where as opposed to the logical operator of equivalence.
In Mathematics it is like equals but A := B means A is defined as B, a triple bar equals can be used to say it's similar and equal by definition but not always the same thing.
Anyway I point to these other references that were probably in the minds of those that invented it, but it's really just that plane equals and less that equals were taken (or potentially easily confused with =<) and something new to define assignment was needed and that made the most sense.
Historical References: I first saw this in SmallTalk the original Object Language, of which SJ of Apple only copied the Windows part of and BG of Microsoft watered down from them further (single threaded). Eventually SJ in NeXT took the second more important lesson from Xerox PARC in, which became Objective C.
Well anyway they just took colon-equals assiment operator from ALGOL 1958 which was later popularized by Pascal
https://en.wikipedia.org/wiki/PARC_(company)
https://en.wikipedia.org/wiki/Assignment_(computer_science)
Assignments typically allow a variable to hold different values at
different times during its life-span and scope. However, some
languages (primarily strictly functional) do not allow that kind of
"destructive" reassignment, as it might imply changes of non-local
state.
The purpose is to enforce referential transparency, i.e. functions
that do not depend on the state of some variable(s), but produce the
same results for a given set of parametric inputs at any point in
time.
https://en.wikipedia.org/wiki/Referential_transparency
For VB.net,
a constructor (for this case, Me = this in Java):
Public ABC(int A, int B, int C){
Me.A = A;
Me.B = B;
Me.C = C;
}
when you create that object:
new ABC(C:=1, A:=2, B:=3)
Then, regardless of the order of the parameters, that ABC object has A=2, B=3, C=1
So, ya, very good practice for others to read your code effectively
Colon-equals was used in Algol and its descendants such as Pascal and Ada because it is as close as ASCII gets to a left-arrow symbol.
The strange convention of using equals for assignment and double-equals for comparison was started with the C language.
In Prolog, there is no distinction between assignment and the equality test.
A feature of Stata that is sometimes inconvenient is that calling a non-defined macro returns the missing value . [edit: Stata actually returns a missing string "", not a numerical missing value], instead of throwing an error.
A piece of code, whose correct execution requires the definition of the macro, may just run giving incorrect results if the macro name is misspelled.
E.g.: having defined
global $options = , vce(robust), when
afterwards one writes reg y x $opt instead of reg y x $options the program runs anyway and it may be difficult to realise that the vce() option was not considered.
Is there any way to force Stata to issue an error in this case or is there some useful trick/best practice that can be used to reduce the risk of incurring this sort of mistake?
The feature is described incorrectly. A macro that is undefined is evaluated as an empty string, conventionally written "", i.e. the delimiters " " contain nothing, or -- if you prefer -- nothing is contained between them.
A macro that is undefined is not ever evaluated as numeric system missing, written as a period . (call it dot or stop if you want).
You would see system missing if the macro were set to contain something else that was system missing, which is entirely different. Saved results from programs, for example, might be system missing.
One way to understand this is that macros in Stata contain strings, not numeric values; the fact that some macros have a numeric interpretation is something else. So, an undefined macro is evaluated as an empty string.
Stata programmers learn to use this feature constructively as a way of allowing defaults when macros are undefined and other choices when they are defined.
You are correct that the feature is a source of bugs, as when a spelling mistake leads Stata to see a name that isn't defined and just ignores the reference. The bug is still the programmer's bug, not Stata's.
So, what can you do, apart from check your code as usual? You can always check whether a macro is defined, as in
if "$options" == "" {
* do something
}
else {
* do something else
}
Conversely,
if "$options" != ""
is a test for content.
Alternatively, you could use string scalars. Here is an experiment:
. sysuse auto, clear
(1978 Automobile Data)
. scalar foo = ", meanonly"
. summarize mpg `=scalar(foo)'
. ret li
scalars:
r(N) = 74
r(sum_w) = 74
r(sum) = 1576
r(mean) = 21.2972972972973
r(min) = 12
r(max) = 41
. summarize mpg `=scalar(bar)'
bar not found
Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
mpg | 74 21.2973 5.785503 12 41
In this case, there was an error message when an undefined scalar was referred to, but the command was executed any way.
Personally, as a long-term (1991- ) and high intensity Stata user, I just use macros routinely and regard being occasionally bitten by bugs of this kind as a very small price to pay for that. I have not ever used string scalars in this sense before trying to answer this question.
It's a different argument, but I regard using global macros in this way as poor programming style. There are general arguments across programming for minimizing the use of globally declared entities. Local macros are the beasts of choice.
I'm a beginner in Prolog and I am dealing with a problem that might seem stupid to you, but I really can't understand what I'm doing wrong! Ok, I have this file fruits.pl and inside that I have something like this:
fruit(apple,small,sweet).
fruit(lemon,small,nosweet).
fruit(melon,big,sweet).
I have already (inside that file made a coexist(X,Y) atom that checks if two fruits can be put together in a plate. It works fine! But now I can't create a suggest(X) that takes as a parameter a fruit and returns a list of fruits that can be put together in the same plate.
The thing is I was trying to make something like that
suggest(X) :- findall(Y,fruit(Y,_,_), List), coexist(X,Y).
What do you think? Every time I try to run this in swi prolog there is a warning 'singleton variable' and when I press
suggest(apple).
then it says false..
sorry for my english :/
Predicates in Prolog do not return anything. You have goals that are satisfied or not and you can interpret that as returning true or false.
Your predicate suggest(X) should contain another parameter that will be bound to the list of fruits that go together with X. An option would be: suggest(X, List) which describes the following relation: List represents all the fruits that go together with X. Then, you could ask:
?- suggest(apple, List).
List = [pear, cherry].
The goal findall(Y, ... , ...) uses the Y variable internally and Y is still unbound after the goal is satisfied. So, you should move coexist(X,Y) inside the second argument of findall/3 which is the goal that is satisfied in all possible ways. Th rule below works only if X is instantiated (suggest(+X, -List)).
suggest(X, List) :- findall(Y, (fruit(Y,_,_), coexist(X, Y)), List).
You can read this as follows: "List represents all fruits Y that coexist with X".
When you try to define a predicate in Prolog, first of all pretend that you have written that predicate already and start with imagining how you would use it. That is, what queries you would like to pose.
To me, it looks as if coexist/2 already describes what you want. BTW, may_coexist/2 might be a more descriptive name. Why do you want this in a separate list? And why using fruit/3 at all? But for the sake of the question let's assume that this makes sense. So essentially you would have now a relation fruit_compatible/2:
fruit_compatible(F, G) :-
fruit(F, _, _),
may_coexist(F, G),
fruit(G, _, _). % maybe you want to add this?
And now, let's assume you want this list too. So you would have a relation fruit_suggestions/2. How to use it?
?- fruit_suggestions(apple, L).
L = [cherry,pear].
or ... should it be rather L = [pear,cherry]? Or both?
?- fruit_suggestions(lemon, L).
L = [orange].
So every time I want a suggestion I have to think of a fruit. Always thinking: what fruit should it be? Fortunately there is a less demanding way in Prolog: Simply use a variable instead of the fruit! Now we should get all suggestions at once!
?- fruit_suggestions(F, L).
F = apple, L = [cherry, pear]
; F = lemon, L = [orange]
; F = cromulon, L = [embiggy, mushfruit].
So we need to implement it such that it will behave that way. findall/3 alone does not solve this. And implementing it manually is far from trivial. But there is setof/3 which handles variables in exactly that manner. Many of the tiny nitty-gritty design decisions have already been made, like that the list will be sorted ascendingly.
fruit_suggestions(F, L) :-
setof(G, fruit_compatible(F, G), L).
Edit: Due to the discussion below, here would be a solution that also permits empty lists. Note that this sounds trivial but it is not. To see this, consider the query:
?- fruit_suggestions(F, []).
What does it mean? What should F be? Also things that are no fruits at all? In that case we would have to produce solutions for everything. Like F = badger ; F = 42 ; .... Most probably this does not make much sense. What might be intended is those fruits that are incompatible with everything. To this end, we need to add a new rule:
fruit_suggestions(F, []) :-
setof(t,X^Y^fruit(F,X,Y),_),
\+ fruit_compatible(F, _).
fruit_suggestions(F, L) :-
setof(G, fruit_compatible(F, G), L).
hi
there is this question in the book that said
Given this grammer
A --> AA | (A) | epsilon
a- what it generates\
b- show that is ambiguous
now the answers that i think of is
a- adjecent paranthesis
b- it generates diffrent parse tree so its abmbiguous and i did a draw showing two scenarios .
is this right or there is a better answer ?
a is almost correct.
Grammar really generates (), ()(), ()()(), … sequences.
But due to second rule it can generate (()), ()((())), etc.
b is not correct.
This grammar is ambiguous due ot immediate left recursion: A → AA.
How to avoid left recursion: one, two.
a) Nearly right...
This grammar generates exactly the set of strings composed of balanced parenthesis. To see why is that so, let's try to make a quick demonstration.
First: Everything that goes out of your grammar is a balanced parenthesis string. Why?, simple induction:
Epsilon is a balanced (empty) parenthesis string.
if A is a balanced parenthesis string, the (A) is also balanced.
if A1 and A2 are balanced, so is A1A2 (I'm using too different identifiers just to make explicit the fact that A -> AA doesn't necessary produces the same for each A).
Second: Every set of balanced string is produced by your grammar. Let's do it by induction on the size of the string.
If the string is zero-sized, it must be Epsilon.
If not, then being N the size of the string and M the length of the shortest prefix that is balanced (note that the rest of the string is also balanced):
If M = N then you can produce that string with (A).
If M < N the you can produce it with A -> AA, the first M characters with the first A and last N - M with the last A.
In either case, you have to produce a string shorter than N characters, so by induction you can do that. QED.
For example: (()())(())
We can generate this string using exactly the idea of the demonstration.
A -> AA -> (A)A -> (AA)A -> ((A)(A))A -> (()())A -> (()())(A) -> (()())((A)) -> (()())(())
b) Of course left and right recursion is enough to say it's ambiguous, but to see why specially this grammar is ambiguous, follow the same idea for the demonstration:
It is ambiguous because you don't need to take the shortest balanced prefix. You could take the longest balanced (or in general any balanced prefix) that is not the size of the string and the demonstration (and generation) would follow the same process.
Ex: (())()()
You can chose A -> AA and generate with the first A the (()) substring, or the (())() substring.
Yes you are right.
That is what ambigious grammar means.
the problem with mbigious grammars is that if you are writing a compiler, and you want to identify each token in certain line of code (or something like that), then ambigiouity wil inerrupt you in identifying as you will have "two explainations" to that line of code.
It sounds like your approach for part B is correct, showing two independent derivations for the same string in the languages defined by the grammar.
However, I think your answer to part A needs a little work. Clearly you can use the second clause recursively to obtain strings like (((((epsilon))))), but there are other types of derivations possible using the first clause and second clause together.