hi im unsure what # stand for in samlltalk and can't find a definition.
exercise code sample:
('hello','world')== #helloworld
I need to find out if this is true or false. I think its false because "==" stands for the same object which it isn't. But i don't know what the # stand for.
This is the syntax for a symbol.
#foobar
Symbols are similar to strings, except that symbols with the same sequence of characters are also exactly the same object.
Some links where you can find this information:
https://rmod-files.lille.inria.fr/FreeBooks/ByExample/08%20-%20Chapter%206%20-%20Special%20Symbol.pdf
https://en.m.wikipedia.org/wiki/Smalltalk#Syntax
Related
I've recently encountered these two variables in some Velocity code:
$!variable1
!$variable2
I was surprised by the similarity of these so I became suspicious about the correctness of the code and become interested in finding the difference between two.
Is it possible that velocity allows any order of these two symbols or do they have different purpose? Do you know the answer?
#Jr. Here is the guide I followed when doing VM R&D: http://velocity.apache.org/engine/1.7/user-guide.html
Velocity uses the !$ and $! annotations for different things. If you use !$ it will basically be the same as a normal "!" operator, but the $! is used as a basic check to see if the variable is blank and if so it prints it out as an empty string. If your variable is empty or null and you don't use the $! annotation it will print the actual variable name as a string.
I googled and stackoverflowed a lot before I finally found the answer at people.apache.org.
According to that:
It is very easy to confuse the quiet reference notation with the
boolean not-Operator. Using the not-Operator, you use !${foo}, while
the quiet reference notation is $!{foo}. And yes, you will end up
sometimes with !$!{foo}...
Easy after all, shame it didn't struck me immediately. Hope this helps someone.
I am reading a Fortran code but I don't understand the real*8 variable declaration with parentheses like.
real*8::xxx21(2,3)
I know real*8 means but what is the meaning of (2,3), its parentheses?
I have tried the similar one from web but no many information there.
This syntax denotes an array of shape 2 times 3.
I highly suggest you to learn the basics of the language before trying to understand any code.
I saw this question and it got me wondering.
Ignoring the fact that pretty much all languages have to be backwards compatible, is there any reason we cannot use operators as both keywords and functions, depending on if it's immediately followed by a parenthesis? Would it make the grammar harder?
I'm thinking mostly of python, but also C-like languages.
Perl does something very similar to this, and the results are sometimes surprising. You'll find warnings about this in many Perl texts; for example, this one comes from the standard distributed Perl documentation (man perlfunc):
Any function in the list below may be used either with or without parentheses around its arguments. (The syntax descriptions omit the parentheses.) If you use parentheses, the simple but occasionally surprising rule is this: It looks like a function, therefore it is a function, and precedence doesn't matter. Otherwise it's a list operator or unary operator, and precedence does matter. Whitespace between the function and left parenthesis doesn't count, so sometimes you need to be careful:
print 1+2+4; # Prints 7.
print(1+2) + 4; # Prints 3.
print (1+2)+4; # Also prints 3!
print +(1+2)+4; # Prints 7.
print ((1+2)+4); # Prints 7.
An even more surprising case, which often bites newcomers:
print
(a % 7 == 0 || a % 7 == 1) ? "good" : "bad";
will print 0 or 1.
In short, it depends on your theory of parsing. Many people believe that parsing should be precise and predictable, even when that results in surprising parses (as in the Python example in the linked question, or even more famously, C++'s most vexing parse). Others lean towards Perl's "Do What I Mean" philosophy, even though the result -- as above -- is sometimes rather different from what the programmer actually meant.
C, C++ and Python all tend towards the "precise and predictable" philosophy, and they are unlikely to change now.
Depending on the language, not() is not defined. If not() is not defined in some language, you can not use it. Why not() is not defined in some language? Because creator of that language probably had not need this type of language construction. Because it is better to let things be simpler.
I didn't touch Prolog since high-school, and even though I've tried to find the info, it didn't help. Below is the example that has to illustrate my problem:
%% everybody():- [dana, cody, bess, abby].
%% Everybody = [dana, cody, bess, abby].
likes(dana, cody).
hates(bess, dana).
hates(cody, abby).
hates(X, Y):- \+ likes(X, Y).
likes_somebody(_, []):- fail.
likes_somebody(X, [girl | others]):-
likes(X, girl) ; likes_somebody(X, others).
likes_everybody(_, []):- true.
likes_everybody(X, [girl | others]):-
likes(X, girl) , likes_everybody(X, others).
maplist(likes_somebody, [dana, cody, bess, abby], [dana, cody, bess, abby]).
How do I declare everybody to just be the list of girls? The commented lines are those which I've tried, but I've got bizarre error messages back.
This is the tutorial I followed more or less so far. I'm using GProlog, if it makes any difference. Sorry for such a basic question. GProlog's manual doesn't deal with language syntax, but I've certainly looked there. As an aside, I would be grateful for information on where to look for language documentation (as opposed to implementation documentation).
Every variable in Prolog must begin with an uppercase letter. So for starters, you want Everybody, not everybody.
Second problem, variables in Prolog are not assignables. So probably what you want to do is make a fact and use that instead:
everybody([dana, cody, bess, abby]).
Your bottom line of code is actually a fact definition and will attempt to overwrite maplist/3. What you probably want to do is put everything above that line into a file (say, called likes.pl) and then consult it ([likes].). Then you can run a query like this:
?- everybody(Everybody), maplist(likes_somebody, Everybody, Everybody).
This won't work, because likes_somebody/2 processes a list in the second argument. The predicate you have for likes_somebody/2 could be written:
likes_somebody(_, []).
but still won't mean much. It simply unifies anything with the empty list:
?- likes_somebody(chicken_tacos, []).
true.
You really need a predicate to tell you if someone is a girl, like this:
girl(dana).
girl(cody).
girl(bess).
girl(abby).
Then you could do what I think you're trying to do, which is something closer to this:
likes_somebody(X) :- girl(X).
Then the maplist construction would work like this:
everybody(Everybody), maplist(likes_somebody, Everybody).
Which would simply return true. You could simplify and eliminate everybody/1 by instead using findall(Girl, girl(X), Everybody) but it's getting weird.
You're trying to do list processing with likes_everybody/2, but it's broken because girl is literally girl, not a variable, and others is literally others, not a list of some kind that could be the tail of another list.
I think you still have some old ideas you need to cleanse. Read some more, write some more, and your code will start to make a lot more sense.
From Parsekit: how to match individual quote characters?
If you define a parser:
#start = int;
int = /[+-]?[0-9]+/
Unfortunately it isn't going to be parsing any integers prefixed with a "+", unless you include:
#numberState = "+" // at the top.
In the number parse above, the "Symbol" default parser wasn't even mentioned, yet it is still active and overrides user defined parsers.
Okay so with numbers you can still fix it by adding the directive. What if you're trying to create a parser for "++"? I haven't found any directive that can make the following parser work.
#start = plusplus;
plusplus = "++";
The effects of default parsers on the user parser seems so arbitrary. Why can't I parse "++"?
Is it possible to just turn off default Parsers altogether? They seem to get in the way if I'm not doing something common.
Or maybe I've got it all wrong.
EDIT:
I've found a parser that would parse plus plus:
#start = plusplus;
plusplus = plus plus;
plus = "+";
I am guessing the answer is: the literal symbols defined in your parser cannot overlap between default parsers; It must be contained completely by at least once of them.
Developer of ParseKit here.
I have a few responses.
I think you'll find the ParseKit API highly elegant and sensible, the more you learn. Keep in mind that I'm not tooting my own horn by saying that. Although I built ParseKit, I did not design the ParseKit API. Rather, the design of ParseKit is based almost entirely on the designs found in Steven Metsker's Building Parsers In Java. I highly recommend you checkout the book if you want to deeply understand ParseKit. Plus it's a fantastic book about parsing in general.
You're confusing Tokenizer States with Parsers. They are two distinct things, but the details are more complex than I can answer here. Again, I recommend Metsker's book.
In the course of answering your question, I did find a small bug in ParseKit. Thanks! However, it was not affecting your outcome described above as you were not using the correct grammar to get the outcome it seems you were looking for. You'll need to update your source code from The Google Code Project now, or else my advice below will not work for you.
Now to answer your question.
I think you are looking for a grammar which both recognizes ++ as a single multi-char Symbol token and also recognizes numbers with leading + chars as explicitly-positive numbers rather than a + Symbol token followed by a Number token.
The correct grammar I believe you are looking for is something like this:
#symbols = '++'; // declare ++ as a multi-char symbol
#numberState = '+'; // allow explicitly-positive numbers
#start = (Number|Symbol)*;
Input like this:
++ +1 -2 + 3 ++
Will be tokenized like so:
[++, +1, -2, +, 3, ++]++/+1/-2/+/3/++^
Two reminders:
Again, you will need to update your source code now to see this work correctly. I had to fix a bug in this case.
This stuff is tricky, and I recommend reading Metsker's book to fully understand how ParseKit works.