I want to make a macro that does this
mac1!("foo", x)
emits
foo(x)
is it even possible?
No, it's not possible. By the time the macro is expanded, the matching is done on the fact that "foo" is an expression (or a literal). The compiler does not distinguish between an expression like "foo" in your example and 123u8, 1 + 2, foo() or { let f = fs::read("foo.txt"); ... } as all of those are expressions. All the macro-by-example knows is that the first parameter is any kind of valid expression and it can't look deeper into it, because the compiler doesn't know what a "type" or a "value" is at this point.
You can use a procedural macro, which can use a parameter's value to generate new tokens, including identifiers.
Related
I see a code as below in https://github.com/terraform-aws-modules/terraform-aws-efs/blob/master/examples/complete/main.tf#L58
# Mount targets / security group
mount_targets = { for k, v in toset(range(length(local.azs))) :
element(local.azs, k) => { subnet_id = element(module.vpc.private_subnets, k) }
}
I am trying to understand what => means here. Also this command with for loop, element and =>.
Could anyone explain here please?
In this case the => symbol isn't an independent language feature but is instead just one part of the for expression syntax when the result will be a mapping.
A for expression which produces a sequence (a tuple, to be specific) has the following general shape:
[
for KEY_SYMBOL, VALUE_SYMBOL in SOURCE_COLLECTION : RESULT
if CONDITION
]
(The KEY_SYMBOL, portion and the if CONDITION portion are both optional.)
The result is a sequence of values that resulted from evaluating RESULT (an expression) for each element of SOURCE_COLLECTION for which CONDITION (another expression) evaluated to true.
When the result is a sequence we only need to specify one result expression, but when the result is a mapping (specifically an object) we need to specify both the keys and the values, and so the mapping form has that additional portion including the => symbol you're asking about:
{
for KEY_SYMBOL, VALUE_SYMBOL in SOURCE_COLLECTION : KEY_RESULT => VALUE_RESULT
if CONDITION
}
The principle is the same here except that for each source element Terraform will evaluate both KEY_RESULT and VALUE_RESULT in order to produce a key/value pair to insert into the resulting mapping.
The => marker here is just some punctuation so that Terraform can unambiguously recognize where the KEY_RESULT ends and where the VALUE_RESULT begins. It has no special meaning aside from being a delimiter inside a mapping-result for expression. You could think of it as serving a similar purpose as the comma between KEY_SYMBOL and VALUE_SYMBOL; it has no meaning of its own, and is only there to mark the boundary between two clauses of the overall expression.
When I read a for expression out loud, I typically pronounce => as "maps to". So with my example above, I might pronounce it as "for each key and value in source collection, key result maps to value result if the condition is true".
Lambda expressions use the operator symbol =, which reads as "goes to." Input parameters are specified on the operator's left side, and statement/expressions are specified on the right. Generally, lambda expressions are not directly used in query syntax but are often used in method calls. Query expressions may contain method calls.
Lambda expression syntax features are as follows:
It is a function without a name.
There are no modifiers, such as overloads and overrides.
The body of the function should contain an expression, rather than a statement.
May contain a call to a function procedure but cannot contain a call to a subprocedure.
The return statement does not exist.
The value returned by the function is only the value of the expression contained in the function body.
The End function statement does not exist.
The parameters must have specified data types or be inferred.
Does not allow generic parameters.
Does not allow optional and ParamArray parameters.
Lambda expressions provide shorthand for the compiler, allowing it to emit methods assigned to delegates.
The compiler performs automatic type inference on the lambda arguments, which is a key advantage.
Vimscript has something like destructuring assignment for lists. For example:
:let [a,b,c] = [1,2,3]
What if I don't care about the first element but I still want to bind b and c? Is there some "wildcard" that I can use to avoid giving a name to the 1?
A possible solution is to use the black hole register:
let [#_,b,c]=[1,2,3]
It seems that you can use it more than once without problems:
let [#_,#_,c]=[1,2,3]
See :h "_.
Just assign those fields to a variable name that makes it clear you're ignoring those values. The _ variable is a great convention for that.
let [_, b, c] = [1, 2, 3]
Or, to ignore the first two terms:
let [_, _, c] = [1, 2, 3]
Note you can use it more than once in a :let-unpack assignment.
In some languages, such as Python, the _ variable special, in that assignments to it are not stored anywhere. That's not the case in Vimscript, a variable with that name is valid and does exist. But if you're performing this assignment from inside a function (which is likely to be the case), then the variable scope will be limited to that function and the namespace pollution shouldn't be concerning.
Here's a real life example usage from the well known vim-figitive plug-in by tpope:
let [_, to, changes; __] = matchlist(a:line, '^ \(.\{-\}\) \+|\zs \(.*\)$')
Note it uses the ; syntax to ignore all fields from the fourth onwards (that's useful to know when unpacking large lists.) It uses _ to ignore the first field and __ to ignore the rest. (Note that using _ for both would have been fine as well.)
The Vim documentation also makes references to using _ as a name for a variable to be ignored. For an example usage of map():
If you do not use "key" you can use a short name:
call map(myDict, {_, val -> 'item: ' . val})
In Haskell, afaik, there are no statements, just expressions. That is, unlike in an imperative language like Javascript, you cannot simply execute code line after line, i.e.
let a = 1
let b = 2
let c = a + b
print(c)
Instead, everything is an expression and nothing can simply modify state and return nothing (i.e. a statement). On top of that, everything would be wrapped in a function such that, in order to mimic such an action as above, you'd use the monadic do syntax and thereby hide the underlying nested functions.
Is this the same in OCAML/F# or can you just have imperative statements?
This is a bit of a complicated topic. Technically, in ML-style languages, everything is an expression. However, there is some syntactic sugar to make it read more like statements. For example, the sample you gave in F# would be:
let a = 1
let b = 2
let c = a + b
printfn "%d" c
However, the compiler silently turns those "statements" into the following expression for you:
let a = 1 in
let b = 2 in
let c = a + b in
printfn "%d" c
Now, the last line here is going to do IO, and unlike in Haskell, it won't change the type of the expression to IO. The type of the expression here is unit. unit is the F# way of expressing "this function doesn't really have result" in the type system. Of course, if the function doesn't have a result, in a purely functional language it would be pointless to call it. The only reason to call it would be for some side-effect, and since Haskell doesn't allow side-effects, they use the IO monad to encode the fact the function has an IO producing side-effect into the type system.
F# and other ML-based languages do allow side-effects like IO, so they have the unit type to represent functions that only do side-effects, like printing. When designing your application, you will generally want to avoid having unit-returning functions except for things like logging or printing. If you feel so inclined, you can even use F#'s moand-ish feature, Computation Expressions, to encapsulate your side-effects for you.
Not to be picky, but there's no language OCaml/F# :-)
To answer for OCaml: OCaml is not a pure functional language. It supports side effects directly through mutability, I/O, and exceptions. In many cases it treats such constructs as expressions with the value (), the single value of type unit.
Expressions of type unit can appear in a sequence separated by ;:
let s = ref 0 in
while !s < 10 do
Printf.printf "%d\n" !s; (* This has type unit *)
incr s (* This has type unit *)
done (* The while as a whole has type unit *)
Update
More specifically, ; ignores the value of the first expression and returns the value of the second expression. The first expression should have type unit but this isn't absolutely required.
# print_endline "hello"; 44 ;;
hello
- : int = 44
# 43 ; 44 ;;
Warning 10: this expression should have type unit.
- : int = 44
The ; operator is right associative, so you can write a ;-separated sequence of expressions without extra parentheses. It has the value of the last (rightmost) expression.
To answer the question we need to define what is an expression and what is a statement.
Distinction between expressions and statements
In layman terms, an expression is something that evaluates (reduces) to a value. It is basically something, that may occur on the right-hand side of the assignment operator. Contrary, a statement is some directive that doesn't produce directly a value.
For example, in Python, the ternary operator builds expressions, e.g.,
'odd' if x % 2 else 'even'
is an expression, so you can assign it to a variable, print, etc
While the following is a statement:
if x % 2:
'odd'
else:
'even'
It is not reduced to a value by Python, it couldn't be printed, assigned to a value, etc.
So far we were focusing more on the semantical differences between expressions and statements. But for a casual user, they are more noticeable on the syntactic level. I.e., there are places where a statement is expected and places where expressions are expected. For example, you can put a statement to the right of the assignment operator.
OCaml/Reason/Haskell/F# story
In OCaml, Reason, and F# such constructs as if, while, print etc are expressions. They all evaluate to values and can occur on the right-hand side of the assignment operator. So it looks like that there is no distinction between statements and expressions. Indeed, there are no statements in OCaml grammar at all. I believe, that F# and Reason are also not using word statement to exclude confusion. However, there are syntactic forms that are not expressions, for example:
open Core_kernel
it is not an expression, definitely, and
type students = student list
is not an expression.
So what is that? In the OCaml parlance, they are called definitions, and they are syntactic constructs that can appear in the module on the, so called, top-level. For example, in OCaml, there are value definitions, that look like this
let harry = student "Harry"
let larry = student "Larry"
let group = [harry; larry]
Every line above is a definition. And every line contains an expression on the right-hand side of the = symbol. In OCaml there is also a let expression, that has form let <v> = <exp> in <exp> that should not be confused with the top-level let definition.
Roughly the same is true for F# and Reason. It is also true for Haskell, that has a distinction between expressions and declarations. It actually should be true to probably every real-world language (i.e., excluding brainfuck and other toy languages).
Summary
So, all these languages have syntactic forms that are not expressions. They are not called statements per se, but we can treat them as statements. So there is a distinction between statements and expressions. The main difference from common imperative languages is that some well-known statements (e.g., if, while, for) are expressions in OCaml/F#/Reason/Haskell, and this is why people commonly say that there is no distinction between expressions and statements.
I want to add non null items to a List. So I do this:
List<Foo> foos = []
Foo foo = makeFoo()
if (foo)
foos << foo
But is there a way to do it in a single operation (without using findAll after the creation of the list). Like:
foos.addNonNull(makeFoo())
Another alternative is to use a short circuit expression:
foo && foos << foo
The foo variable must evaluate to true for the second part to be evaluated. This is a common practice in some other languages but I'd hesitate to use it widely in groovy due to readability issues and conventions.
No, you'd need to use an if, or write your own addNonNull method (which just uses an if)
Also:
if( foo ) {
probably isn't enough, as this will skip empty strings, or 0 if it returns integers
You'd need
if( foo != null ) {
The answer is YES! we can get rid of assigning a variable
Foo foo = makeFoo()//we can ditch this
The answer is NO we can't get rid of the condition. BUT we can make it more compact.
Here's how
List<Foo> foos = []
foos += (makeFoo()?:[]);
The trick is groovy's "+" operator which works differently based on what is to the left and what is to the right of the "+". It just so happens that if what is on the left is a list and what is on the right is an empty list, nothing gets added to the list on the left.
Pros are it is quick to type and compact.
Cons are it is not instantly obvious what is happening to most people
AND we replaced the variable assignment with an extra operation. Groovy is
going to try to do something to List foos no matter what, it just so happens that in the second case the result of that operation gives us a desired result.
In some dynamic languages I have seen this kind of syntax:
myValue = if (this.IsValidObject)
{
UpdateGraph();
UpdateCount();
this.Name;
}
else
{
Debug.Log (Exceptions.UninitializedObject);
3;
}
Basically being able to return the last statement in a branch as the return value for a variable, not necessarily only for method returns, but they could be achieved as well.
What's the name of this feature?
Can this also be achieved in staticly typed languages such as C#? I know C# has ternary operator, but I mean using if statements, switch statements as shown above.
It is called "conditional-branches-are-expressions" or "death to the statement/expression divide".
See Conditional If Expressions:
Many languages support if expressions, which are similar to if statements, but return a value as a result. Thus, they are true expressions (which evaluate to a value), not statements (which just perform an action).
That is, if (expr) { ... } is an expression (could possible be an expression or a statement depending upon context) in the language grammar just as ?: is an expression in languages like C, C# or Java.
This form is common in functional programming languages (which eschew side-effects) -- however, it is not "functional programming" per se and exists in other language that accept/allow a "functional like syntax" while still utilizing heavy side-effects and other paradigms (e.g. Ruby).
Some languages like Perl allow this behavior to be simulated. That is, $x = eval { if (true) { "hello world!" } else { "goodbye" } }; print $x will display "hello world!" because the eval expression evaluates to the last value evaluated inside even though the if grammar production itself is not an expression. ($x = if ... is a syntax error in Perl).
Happy coding.
To answer your other question:
Can this also be achieved in staticly typed languages such as C#?
Is it a thing the language supports? No. Can it be achieved? Kind of.
C# --like C++, Java, and all that ilk-- has expressions and statements. Statements, like if-then and switch-case, don't return values and there fore can't be used as expressions. Also, as a slight aside, your example assigns myValue to either a string or an integer, which C# can't do because it is strongly typed. You'd either have to use object myValue and then accept the casting and boxing costs, use var myValue (which is still static typed, just inferred), or some other bizarre cleverness.
Anyway, so if if-then is a statement, how do you do that in C#? You'd have to build a method to accomplish the goal of if-then-else. You could use a static method as an extension to bools, to model the Smalltalk way of doing it:
public static T IfTrue(this bool value, Action doThen, Action doElse )
{
if(value)
return doThen();
else
return doElse();
}
To use this, you'd do something like
var myVal = (6 < 7).IfTrue(() => return "Less than", () => return "Greater than");
Disclaimer: I tested none of that, so it may not quite work due to typos, but I think the principle is correct.
The new IfTrue() function checks the boolean it is attached to and executes one of two delegates passed into it. They must have the same return type, and neither accepts arguments (use closures, so it won't matter).
Now, should you do that? No, almost certainly not. Its not the proper C# way of doing things so it's confusing, and its much less efficient than using an if-then. You're trading off something like 1 IL instruction for a complex mess of classes and method calls that .NET will build behind the scenes to support that.
It is a ternary conditional.
In C you can use, for example:
printf("Debug? %s\n", debug?"yes":"no");
Edited:
A compound statement list can be evaluated as a expression in C. The last statement should be a expression and the whole compound statement surrounded by braces.
For example:
#include <stdio.h>
int main(void)
{
int a=0, b=1;
a=({
printf("testing compound statement\n");
if(b==a)
printf("equals\n");
b+1;
});
printf("a=%d\n", a);
return 0;
}
So the name of the characteristic you are doing is assigning to a (local) variable a compound statement. Now I think this helps you a little bit more. For more, please visit this source:
http://www.chemie.fu-berlin.de/chemnet/use/info/gcc/gcc_8.html
Take care,
Beco.
PS. This example makes more sense in the context of your question:
a=({
int c;
if(b==a)
c=b+1;
else
c=a-1;
c;
});
In addition to returning the value of the last expression in a branch, it's likely (depending on the language) that myValue is being assigned to an anonymous function -- or in Smalltalk / Ruby, code blocks:
A block of code (an anonymous function) can be expressed as a literal value (which is an object, since all values are objects.)
In this case, since myValue is actually pointing to a function that gets invoked only when myValue is used, the language probably implements them as closures, which are originally a feature of functional languages.
Because closures are first-class functions with free variables, closures exist in C#. However, the implicit return does not occur; in C# they're simply anonymous delegates! Consider:
Func<Object> myValue = delegate()
{
if (this.IsValidObject)
{
UpdateGraph();
UpdateCount();
return this.Name;
}
else
{
Debug.Log (Exceptions.UninitializedObject);
return 3;
}
};
This can also be done in C# using lambda expressions:
Func<Object> myValue = () =>
{
if (this.IsValidObject) { ... }
else { ... }
};
I realize your question is asking about the implicit return value, but I am trying to illustrate that there is more than just "conditional branches are expressions" going on here.
Can this also be achieved in staticly
typed languages?
Sure, the types of the involved expressions can be statically and strictly checked. There seems to be nothing dependent on dynamic typing in the "if-as-expression" approach.
For example, Haskell--a strict statically typed language with a rich system of types:
$ ghci
Prelude> let x = if True then "a" else "b" in x
"a"
(the example expression could be simpler, I just wanted to reflect the assignment from your question, but the expression to demonstrate the feature could be simlpler:
Prelude> if True then "a" else "b"
"a"
.)