in this code
void something () { /*something*/ }
are () separators or operators?
as i know in a function call () are operators:
something();
but in a function definition it would be a bit weird to have an operator, because operator in fact is a function and there is a function in a function definition?
can somebody explain this topic? what are the separators/punctuators exactly? they are tokens for the compiler to differentiate some part of the code? for example two statements?
/*statement1*/;
/*statement2*/;
;s are separating the statements from each other
so they are atomic syntactic elements for the compiler to understand the source code?
Depends on the context.
() in C++ can fulfill both definitions at once (operator and separator), or only one at a time.
It is an operator, since () is literally defined as the function call operator in the language spec. Since an overloaded operator is still an operator, this is independent of the number of arguments being passed to it (zero, or several).
A separator in terms of (programming) languages is usually defined as one or two tokens that separate some language features from other language features. This is the case when you pass parameters to a function when it is called, since the brackets separate the function name from the function arguments. This is not the case if no argument argument is being passed during the function call, as there is nothing to separate. In this case, () would act as an operator, but not as a separator.
I also almost forgot to mention the fact that round brackets are also used in arithmetic to denote precedence (acting as separators, not operators).
Another example of () acting as operator, but not as a separator, would be a cast.
Related
I'm writing a vscode extension for my programming language. I'm having difficulty coming up with a tmlanguage rule to parse my function declarations.
A function can look like this:
def macro_1 macro_2(foo, 20) function1(a: int, b: double) -> int {
}
macro_1 and macro_2 would be keywords while function1 should be a function name
The parser is looking for matching parenthesis and then looking ahead if it is followed by an identifier, in which case it is assumed to be a macro invocation. There can be any amount of macro invocations on a function. Do note that lines can be continued with a backslash, but I'm not trying to implement that yet.
My first idea was to use a single rule to parse both macro invocations and the function name with parameters, but the problem with this is that the macro invocation can have any expression as an argument while a function parameter can only be an identifier followed by colon and a type.
Are there advantages in defining function arguments as keywords rather than just normal arguments? Was hoping to find the answer on http://docs.racket-lang.org/reference/define.html
Use keywords arguments when
- there are a lot of arguments
- most arguments have default arguments
For a function with, say, 10 arguments it can be difficult to remember the order of the argument. When keywords are used, then the order doesn't matter.
I was looking at the source for the r_str macro in Julia, which parses r"text" into Regex("text"). The second argument is flags..., which passes flags into the regex, like i for case insensitive, and so on.
I was playing with this myself and got:
julia> macro a_str(p, flags...)
print(flags)
p
end
julia> a"abc"iii
("iii",)"abc"
So it seems that the iii is all passed in as the first flag. In that case, why is there the ... on the flags. Is it possible to pass in more than one element of flags to the macro?
When this question was originally asked, a macro expander – i.e. the function defined with the macro keyword, which is called to transform the expressions passed to a macro into a single output expression – was not a generic function, but rather an anonymous function, which were a different kind of function in Julia 0.4 and earlier. At that point, the only way to write an anonymous function signature which could work for either one or two arguments was to use a trailing varargs argument, which is why this pattern was used to define string macros. In Julia 0.5 all functions have become generic functions, including anonymous functions and macro expanders. Thus, you can now write a macro a variety of ways, including the old way of using a varargs argument after the string argument:
# old style
macro rm_str(raw, rest...)
remove = isempty(rest) ? "aeiouy" : rest[1]
replace(raw, collect(remove), "")
end
# new style with two methods
macro rm_str(raw)
replace(raw, ['a','e','i','o','u','y'], "")
end
macro rm_str(raw, remove)
replace(raw, collect(remove), "")
end
# new style with default second argument
macro rm_str(raw, remove="aeiouy")
replace(raw, collect(remove), "")
end
These all result in the same non-standard string literal behavior:
julia> rm"foo bar baz"
"f br bz"
julia> rm"foo bar baz"abc
"foo r z"
The string literal produces the string with the flagged letters stripped from it, defaulting to stripping out all the ASCII vowels ("aeiouy"). The new approach of using a second argument with a default is the easiest and clearest in this case, as it will be in many cases, but now you can use whichever approach is best for the circumstances.
With an explicit call like
#a_str("abc", "iii", "jjj")
you can pass multiple flags. But I'm not aware of a way to make this work with a"abc"ijk syntax.
I don't believe it is possible, and the documentation doesn't provide an example where that would be used. In addition, the mostly-fully-compliant JuliaParser.jl doesn't support multiple flags either. Perhaps open an PR on Julia changing that?
The difference between the two is not so clear from the Cadence documentation.
Could someone please elaborate on the difference between the two?
A define as macro is just a plain old macro that you probably know from other programming languages. It just means that at some select locations in the macro code you can substitute your own code.
A define as computed macro allows you to construct your output code programmatically, by using control flow statements (if, for, etc.). It acts kind of like a function that returns a string, with the return value being the code that will be inserted in its place by the pre-processor.
With both define as and define as computed macros you define a new syntactic construct of a given syntactic category (for example, <statement> or <action>), and you implement the replacement code that replaces a construct matching the macro match expression (or pattern).
In both cases the macro match expression can have syntactic arguments that are used inside the replacement code and are substituted with the actual code strings used in the matched code.
The difference is that with a define as macro the replacement code is just written in the macro body.
With a define as computed macro you write a procedural code that computes the desired replacement code text and returns it as a string. It's effectively a method that returns string, you can even use the result keyword to assign the resulting string, just like in any e method.
A define as computed macro is useful when the replacement code is not fixed, and can be different depending on the exact macro argument values or even semantic context (for example, in some cases a reflection query can be used to decide on the exact replacement code).
(But it's important to remember that even define as computed macros are executed during compilation and not at run time, so they cannot query actual run time values of fields or variables to decide on the resulting replacement code).
Here are some important differences between the two macro kinds.
A define as macro is more readable and usually easier to write. You just write down the code that you want to be created.
Define as computed macros are stronger. Everything that can be implemented with define as, can also be implemented with define as computed, but not vice versa. When the replacement code is not fixed, define as is not sufficient.
A define as macro can be used immediately after its definition. If the construct it introduces is used in the statement just following the macro, it will already be matched. A define as computed macro can only be used in the next file, and is not usable in the same file in which the macro is defined.
I know that my is used to declare a variable local to a block or file. I have always assumed that my is a keyword in Perl. But I was just told that it's actually a function. One of the proofs is that perldoc puts my under the “Functions” section, see http://perldoc.perl.org/functions/my.html.
How does a function do the job of declaring local variables?
my is not a function, it's just clumped together with functions (in perl documentation) because it works like a function.
If you look at perldoc perlfunc, it is saith,
Here are Perl's functions (including things that look like functions, like some keywords and named operators) arranged by category...
then a bit below that
Keywords related to scoping
caller, import, local, my, our, package, state, use
Specifically, note that the word “keyword” was used there instead of “function”
So that implies that you would find some non-functions (e.g. keywords) under Perl functions A-Z
Another way of saying this: if something is listed under “Functions” in perldoc, it is not necessarily a function – it can be a keyword or named operator which acts like a function.
Yes, by Perl's (very unique) definition, my is a function. The opening paragraph of perlfunc defines "function":
The functions in this section can serve as terms in an expression. They fall into two major categories: list operators and named unary operators.
my is a named operator. But it's special in two ways:
In addition to behaving like a function (that allocates a new variable and returns that variable), it has a compile-time effect.
my ... is a unary operator, but it can accept multiple arguments when parens are used.
If on the other hand you were ask if my was a function by C's definition, then no. my is not a C function. Neither is print, open, chr, etc. Everything in perlfunc is an operator; none of them are functions.
Finally, print, open and chr are far closer to a person's conception of a function than my. To be more precise, few people would consider my to be a function. It's more of a technicality than anything meaningful that it matches perfunc's definition of function.
See also:
What are perl built-in operators/functions?
Why does this [my] variable keep its value