I believe most operators will not modify their operands. If I want to ask if an operator is one of those that will alter it's operands, is there a word I can use that means that? For example:
Is the bitshift operator a [adjective that means "it will change the operand"]
or
The '+' operator will not modify a variable, but changing it to '+=' will change the varable on the left, because '+=' is a [need that word again, this time in noun form]
Generally, unary operators and some shorthand operators are those which are used to change the operand.
++a;
--a;
a++;
k +=30;
k -=30;
etc.
Related
Hi may i know what does the below query means?
REGEXP_REPLACE(number,'[^'' ''-/0-9:-#A-Z''[''-`a-z{-~]', 'xy') ext_number
part 1
In terms of explaining what the function function call is doing:
It is a function call to analyse an input string 'number' with a regex (2nd argument) and replace any parts of the string which match a specific string. As for the name after the parenthesis I am not sure, but the documentation for the function is here
part 2
Sorry to be writing a question within an answer here but I cannot respond in comments yet (not enough rep)
Does this regex work? Unless sql uses different syntax this would appear to be a non-functional regex. There are some red flags, e.g:
The entire regex is wrapped in square parenthesis, indicating a set of characters but seems to predominantly hold an expression
There is a range indicator between a single quote and a character (invalid range: if a dash was required in the match it should be escaped with a '\' (backslash))
One set of square brackets is never closed
After some minor tweaks this regex is valid syntax:
^'' ''\-\/0-9:-#A-Z''[''-a-z{-~]`, but does not match anything I can think of, it is important to know what string is being examined/what the context is for the program in order to identify what the regex might be attempting to do
It seems like it is meant to replaces all ASCII control characters in the column or variable number with xy.
[] encloses a class of characters. Any character in that class matches. [^] negates that, hence all characters match, that are not in the class.
- is a range operator, e.g. a-z means all characters from a to z, like abc...xyz.
It seams like characters enclosed in ' should be escaped (The second ' is to escape the ' in the string itself.) At least this would make some sense. (But for none of the DBMS I found having a regexp_replace() function (Postgres, Oracle, DB2, MariaDB, MySQL), I found something in the docs, that would indicate this escape mechanism. They all use \, but maybe I missed something? Unfortunately you didn't tag which DBMS you're actually using!)
Now if you take an ASCII table you'll see, that the ranges in the expression make up all printable characters (counting space as printable) in groups from space to /, 0 to 9, : to #, etc.. Actually it might have been shorter to express it as '' ''-~, space to ~.
Given the negation, all these don't match. The ones left are from NUL to US and DEL. These match and get replaced by xy one by one.
In Backus–Naur Form would:
print_stmt : "print" (string | expr)+
match to:
print string
print expr
or
print (string)
print (expr)
I'm not sure whether the parentheses have to be there or not.
It would match either of the first two options, and a number of other possibilities.
In this dialect of BNF, it appears that the parentheses are metacharacters. The + probably means 'one or more' of the previous units, but if the ) was repeated one or more times, it would be a very unusual language. If the + was absent, then either interpretation would be reasonable and I couldn't give as confident an answer; you would have to go back and find the specification for the dialect of BNF you're interpreting.
Because of the +, this should also be valid:
print string string expr string
I'm writing a regular expression in Objective-C.
The escape sequence \w is illegal and emits a warning, so the regular expression /\w/ must be written as #"\\w"; the escape sequence \? is valid, apparently, and doesn't emit a warning, so the regular expression /\?/ must be written as #"\\?" (i.e., the backslash must be escaped).
Question marks aren't invisible like \t or \n, so why is \? a valid escape sequence?
Edit: To clarify, I'm not asking about the quantifier, I'm asking about a string escape sequence. That is, this doesn't emit a warning:
NSString *valid = #"\?";
By contrast, this does emit a warning ("Unknown escape sequence '\w'"):
NSString *invalid = #"\w";
It specifies a literal question mark. It is needed because of a little-known feature called trigraphs, where you can write a three-character sequence starting with question marks to substitute another character. If you have trigraphs enabled, in order to write "??" in a string, you need to write it as "?\?" in order to prevent the preprocessor from trying to read it as the beginning of a trigraph.
(If you're wondering "Why would anybody introduce a feature like this?": Some keyboards or character sets didn't include commonly used symbols like {. so they introduced trigraphs so you could write ??< instead.)
? in regex is a quantifier, it means 0 or 1 occurences. When appended to the + or * quantifiers, it makes it "lazy".
For example, applying the regex o? to the string foo? would match o.
However, the regex o\? in foo? would match o?, because it is searching for a literal question mark in the string, instead of an arbitrary quantifier.
Applying the regex o*? to foo? would match oo.
More info on quantifiers here.
I'm looking at inherited code and I found this in a vb.net windows form:
New System.Drawing.SizeF(6.0!, 13.0!)
My question is, what is the significance of the ! (exclamation) operator here? Most of my searching for the exclamation operator ends up returning recordset format or the ! gets ignored in the search and I get hundreds of unrelated items.
It's to force the literal to be a Single.
Visual Basic supports Type Characters:
In addition to specifying a data type in a declaration statement, you can force the data type of some programming elements with a type character. The type character must immediately follow the element, with no intervening characters of any kind.
and:
Literals can also use the identifier type characters (%, &, #, !, #, $), as can variables, constants, and expressions. However, the literal type characters (S, I, L, D, F, R, C) can be used only with literals.
In this case, the ! stands for Single:
Type Characters. Appending the literal type character F to a literal forces it to the Single data type. Appending the identifier type character ! to any identifier forces it to Single.
(emphasis mine)
It is a type character. It means that 6.0 is a Single data type.
http://msdn.microsoft.com/en-us/library/s9cz43ek.aspx show the type characters.
I defined a custom equality operator (the definition is not really important so I will insert dummy stuff):
let ( ~=~ ) a b = true
If I try to use it infix:
if a ~=~ b then 1 else 2
I get the following error:This expression is not a function; it cannot be applied.
I can fix this either by renaming the operator from ~=~ to =~ or by calling it as a function: if (~=~) a b then 1 else 2.
This seems that is a general problem with operators that start with ~.
My question is why I can't use such operators infix? Is anything special about ~ symbol?
Note: I already went through documentation but I couldn't find anything relevant. Maybe I missed something?
In OCaml, whether an operator is infix or prefix is determined by its first character.
In you case, the character '~' is for prefix: by let (~=~) a b = ..., you are defining a prefix operator. ~=~ a is a valid expression, and returns a function.
In addition to infix or prefix, infix operator associativity (left or right) and operator precedences (which of + and * has stronger?) are syntactically determined by the first character of the operator.
This sounds ugly, since you cannot have control of your fancy operators characteristics, but it makes easier to read OCaml source code by someone else with lots of strange custom operators.
Here is the table of chars for operators:
The first char : prefix/infix/connectivity power/left-or-right
! ~ ? : prefix
= < > | & $ : infix0, left
# ^ : infix1, right
+ - : infix2, left
* / : infix3, left ( ** is exceptional. It is right assoc and have power 4)
By lexical conventions of ocaml ~ is reserved for prefix operators, see
http://caml.inria.fr/pub/docs/manual-ocaml/lex.html#infix-symbol