The following code got compilation error:
var a : Int = 0
var b : Int = 3
var sum : Int = 0
while (sum = a+b) < 2 {
}
The error message is:
Cannot invoke '<' with an argument list of type '((()),
IntegerLiteralConvertible)'
How to solve this problem? (Of course I can put sum assignment statement out side the while statement. But this is not convenient. Any other advice? Thanks
In many other languages, including C and Objective-C, sum = a+b would return the value of sum, so it could be compared.
In Swift, this doesn't work. This was done intentionally to avoid a common programmer error. From The Swift Programming Language:
Swift supports most standard C operators and improves several capabilities to eliminate common coding errors. The assignment operator (=) does not return a value, to prevent it from being mistakenly used when the equal to operator (==) is intended.
Since the assignment operator does not return a value, it can't be compared with another value.
It is not possible to overload the default assignment operator (=), but you could create a new operator or overload one of the compound operators to add this functionality. However, this would be unintuitive to future readers of your code, so you may want to simply move the assignment to a separate line.
In most languages, assignments propagate their value -- that is, when you call
sum = a + b
the new value of sum is available for another part of the expression:
doubleSum = (sum = a + b) * 2
Swift doesn't work that way -- the value of sum isn't available after the assignment, so it can't be compared in your while statement. From Apple's documentation:
This feature prevents the assignment operator (=) from being used by
accident when the equal to operator (==) is actually intended. By
making if x = y invalid, Swift helps you to avoid these kinds of
errors in your code.
The other answers explain why your code won't compile. Here is how you can clean it up without calculating sum in the while loop (I'm assuming you want to be able to reassign what sum's getter is, elsewhere.):
var a = 0, b = 3
var getSum = { a + b }
var sum: Int { return getSum() }
while sum < 2 {
...and if you're okay with invoking sum with parentheses:
var a = 0, b = 3
var sum = { a + b }
while sum() < 2 {
You can rewrite it as a for loop, although you'll have to repeat the assignment and addition:
for sum = a+b; sum < 2; sum = a+b {
}
Related
i'm trying to develop the algorithm W in f# for type inference, but i would like to understand how to write the function for generating fresh variables properly.
Actually my function is
let counter = ref -1
let generate_fresh_variable () : string =
let list_variables = ['a' .. 'z'] |> List.map string
counter.Value <- !counter + 1
list_variables.Item(!counter)
but i'm not satisfy with this solution, someone can give me other better ideas?
If you really want to do this with an impure function, I would write it like this:
let mutable counter = -1
let generate_fresh_variable () =
counter <- counter + 1
counter + int 'a'
|> char
|> string
Notes:
Reference cells are obsolete. If you need impurity, use mutable variables instead. (Alternatively, if you really want to stick with a reference cell, the canonical way to update it is with :=, rather than assigning directly to the underlying Value.)
There's no need to maintain a list of potential variable names (and there's especially no need to rebuild the entire list each time you generate a fresh variable).
What happens if you need more than 26 variables?
If you wanted to use some more sophisticated F# tricks, you could create an inifinte sequence of names using a sequence expression (which makes it very easy to handle the looping and dealing with >26 names):
let names = seq {
for i in Seq.initInfinite id do
for c in 'a' .. 'z' do
if i = 0 then yield string c
else yield string c + string i }
A function to get the fresh name would then pick the next name from the sequence. You need to do this using the underlying enumerator. Another nice trick is to hide the state in a local variable and return a function using lambda:
let freshName =
let en = names.GetEnumerator()
fun () ->
ignore(en.MoveNext())
en.Current
Then just call freshName() as many times as you need.
I want to use a variable that references an arithmetic operator within an if statement expression as shown below:
str = { '>60', '>60', '>-60', '=0' }
del = 75
function decode_prog(var1, var2)
op = string.sub(var1, 1, 1)
vb = tonumber(string.sub(var1, 2, 3))
if var2 op vb then
print("condition met")
else
print('condition not meet')
end
end
for i = 1, #str do
decode_prog(str[i], del)
end
When the above code executes, it should either print "condition met" or "condition not met" based on the result of the operation, however I am instead receiving an error.
You cannot substitute a native Lua operator with a variable that references a function, the only way to go about what you are attempted to do is to create a set of functions within an associative array and set the index as a reference to the respective operation you want to conduct.
Looking at your list, you have a greater than (>) and equal to (=). We create a table for these operations that takes two parameters as follows.
local operators = {
[">"] = function(x, y) return x > y end,
["="] = function(x, y) return x == y end,
-- Add more operations as required.
}
You can then invoke the respective function from the decode_prog function by obtaining the operation character from the string, along with the numeric value itself - this is possible because you can obtain the function from the associative array where the index is the string of the operation we want to conduct.
local result = operators[op](var2, number)
This calls upon the operators array, uses the op to determine which index we need to go to for our appropriate operation, and returns the value.
Final Code:
str = { '>60', '>60', '>-60', '=0' }
del = 75
local operators = {
[">"] = function(x, y) return x > y end,
["="] = function(x, y) return x == y end,
}
function decode_prog(var1, var2)
local op = string.sub(var1, 1, 1) -- Fetch the arithmetic operator we intend to use.
local number = tonumber(string.sub(var1, 2)) -- Strip the operator from the number string and convert the result to a numeric value.
local result = operators[op](var2, number) -- Invoke the respective function from the operators table based on what character we see at position one.
if result then
print("condition met")
else
print('condition not meet')
end
end
for i = 1, #str do
decode_prog(str[i], del)
end
I can't make much sense of your code or what you want to achieve doing that but if could simply use load.
You build your expression as a string and run it. Of course you should take care of two character operators like >= which I did not and you should validate your input.
local str={'>60','>60','>-60','=0'}
local del=75
function decode_prog(var1, var2)
local operator = var1:sub(1,1):gsub("=", "==")
local expr = string.format("return %d %s %s", var2,operator, var1:sub(2))
print(string.format("condition %smet", load(expr)() and "" or "not "))
end
for i,v in ipairs(str) do
decode_prog(v, del)
end
A very simple way would be to add a condition for each supported operator:
function decode_prog(var1, var2)
op = string.sub(var1, 1, 1)
vb = tonumber(string.sub(var1, 2)) --remove the last argument and use tonumber()
if vb == nil then return end --if the string does not contain number
if (op == ">" and var2 > vb) or (op == "=" and var2 == vb) --[[add more conditions here]] then
print("condition met")
else
print("condition not met")
end
end
I changed the vb=string.sub(var1,2,3) line too.
This form vb = tonumber(string.sub(var1, 2)) will allow use of numbers that have any number of digits and added tonumber() which will allow us to catch not-a-number errors when comparison would probably fail.
Then I added a logic to determine what the operator is and if the condition is met.
Operator limitations:
This will only work with operators that are one character and operator such as >= will not be possible unless you use a different character for it. ≥ will not play nicely, since it is multiple characters.
I am new to Dart and I can see Dart has num which is the superclass of int and double (and only those two, since it's a compile time error to subclass num to anything else).
So far I can't see any benefits of using num instead of int or double, are there any cases where num is better or even recommended? Or is it just to avoid thinking about the type of the number so the compiler will decide if the number is an int or a double for us?
One benefit for example
before Dart 2.1 :
suppose you need to define a double var like,
double x ;
if you define your x to be a double, when you assign it to its value, you have to specify it say for example 9.876.
x = 9.876;
so far so good.
Now you need to assign it a value like say 9
you can't code it like this
x = 9; //this will be error before dart 2.1
so you need to code it like
x = 9.0;
but if you define x as num
you can use
x = 9.0;
and
x = 9;
so it is a convenient way to avoid these type mismatch errors between integer and double types in dart.
both will be valid.
this was a situation before Dart 2.1 but still can help explain the concept
check this may be related
Not sure if this is useful to anyone, but I just ran into a case where I needed num in a way.
I defined a utility function like this:
T maximumByOrNull<T, K extends Comparable<K>>(Iterable<T> it, K key(T),) {
return it.isEmpty
? null
: it.reduce((a, b) => key(a).compareTo(key(b)) > 0 ? a : b);
}
And invoking it like this…
eldest = maximumByOrNull(students, (s) => s.age);
… caused trouble when age is an int, because int itself does not implement Comparable<int>.
So Dart cannot infer the type K in the invocation to maximumByOrNull.
However, num does implement Comparable<num>. And if I specified:
eldest = maximumByOrNull(students, (s) => s.age as num); // this
eldest = maximumByOrNull<Student, num>(students, (s) => s.age); // or this
the complaint went away.
Bottom line: it seems num implements Comparable when int and double do not, themselves, and sometimes this causes trouble for generic functions.
A good use case of num are extensions that work with int and double.
As an example I include the extension MinMax on List<num> that provides the getters min and max.
extension MinMax on List<num>{
/// Returns the maximum value or `null` if the list is empty.
num get max {
return (isNotEmpty)
? fold<num>(0, (prev, current) => (prev > current) ? prev : current)
: null;
}
/// Returns the minimum value or `null` if the list is empty.
num get min {
return (isNotEmpty)
? fold<num>(0, (prev, current) => (prev < current) ? prev : current)
: null;
}
}
Using the extension above one can access the min/max values without a need to create specific implementations for the classes int and double.
void main() {
final a = <int>[1,3,5];
final b = <double>[ 0.5, 0.8, -5.0];
print(a.min);
print(b.max);
}
I just ran into a use case when it is useful.
My app stores weight, which was originally defined as a double.
When using with a local database (sqlite) it works fine, since sqlite handles integer and real types.
However, when I converted my app to use Firestore database, I ran into issues with all my double fields. If a decimal value is stored everything works fine. However, when the weight happens to be a whole number, Firestore returns it as an int and suddenly type errors - int is not a subtype of type double - start to appear.
In the above scenario changing the double fields and variables to num was a quite simple solution.
let i = ref 123
let j = ref 123
i = j // true
Similarly:
let i = box 123
let j = box 123
i = j // true
Presumably, i and j are not actually pointing to the same exact place in memory...??
I get around this (odd?) behavior in the second case by doing:
obj.ReferenceEquals (i, j) // false
What is the proper equality test for the first case?
EDIT:
I see that calling obj.ReferenceEquals works in the first case, as well.
Can someone out there explain to me why I have to call this function, though? Why can't I just use the = operator?
The (=) operator calls GenericEqualityObj. It first checks the types of the args (for arrays, assignability to IStructuralEquatable, and a few other special cases) and the final case calls obj.Equals. Equals is overridden by ValueType (which int derives) to do bit comparison. That explains why (box 123) = (box 123) is true.
Ref cells are represented using records, by default records are structurally comparable\equatable
I wonder what this means in F#.
“a function taking an integer, which returns a function which takes an integer and returns an integer.”
But I don't understand this well.
Can anyone explain this so clear ?
[Update]:
> let f1 x y = x+y ;;
val f1 : int -> int -> int
What this mean ?
F# types
Let's begin from the beginning.
F# uses the colon (:) notation to indicate types of things. Let's say you define a value of type int:
let myNumber = 5
F# Interactive will understand that myNumber is an integer, and will tell you this by:
myNumber : int
which is read as
myNumber is of type int
F# functional types
So far so good. Let's introduce something else, functional types. A functional type is simply the type of a function. F# uses -> to denote a functional type. This arrow symbolizes that what is written on its left-hand side is transformed into what is written into its right-hand side.
Let's consider a simple function, that takes one argument and transforms it into one output. An example of such a function would be:
isEven : int -> bool
This introduces the name of the function (on the left of the :), and its type. This line can be read in English as:
isEven is of type function that transforms an int into a bool.
Note that to correctly interpret what is being said, you should make a short pause just after the part "is of type", and then read the rest of the sentence at once, without pausing.
In F# functions are values
In F#, functions are (almost) no more special than ordinary types. They are things that you can pass around to functions, return from functions, just like bools, ints or strings.
So if you have:
myNumber : int
isEven : int -> bool
You should consider int and int -> bool as two entities of the same kind: types. Here, myNumber is a value of type int, and isEven is a value of type int -> bool (this is what I'm trying to symbolize when I talk about the short pause above).
Function application
Values of types that contain -> happens to be also called functions, and have special powers: you can apply a function to a value. So, for example,
isEven myNumber
means that you are applying the function called isEven to the value myNumber. As you can expect by inspecting the type of isEven, it will return a boolean value. If you have correctly implemented isEven, it would obviously return false.
A function that returns a value of a functional type
Let's define a generic function to determine is an integer is multiple of some other integer. We can imagine that our function's type will be (the parenthesis are here to help you understand, they might or might not be present, they have a special meaning):
isMultipleOf : int -> (int -> bool)
As you can guess, this is read as:
isMultipleOf is of type (PAUSE) function that transforms an int into (PAUSE) function that transforms an int into a bool.
(here the (PAUSE) denote the pauses when reading out loud).
We will define this function later. Before that, let's see how we can use it:
let isEven = isMultipleOf 2
F# interactive would answer:
isEven : int -> bool
which is read as
isEven is of type int -> bool
Here, isEven has type int -> bool, since we have just given the value 2 (int) to isMultipleOf, which, as we have already seen, transforms an int into an int -> bool.
We can view this function isMultipleOf as a sort of function creator.
Definition of isMultipleOf
So now let's define this mystical function-creating function.
let isMultipleOf n x =
(x % n) = 0
Easy, huh?
If you type this into F# Interactive, it will answer:
isMultipleOf : int -> int -> bool
Where are the parenthesis?
Note that there are no parenthesis. This is not particularly important for you now. Just remember that the arrows are right associative. That is, if you have
a -> b -> c
you should interpret it as
a -> (b -> c)
The right in right associative means that you should interpret as if there were parenthesis around the rightmost operator. So:
a -> b -> c -> d
should be interpreted as
a -> (b -> (c -> d))
Usages of isMultipleOf
So, as you have seen, we can use isMultipleOf to create new functions:
let isEven = isMultipleOf 2
let isOdd = not << isEven
let isMultipleOfThree = isMultipleOf 3
let endsWithZero = isMultipleOf 10
F# Interactive would respond:
isEven : int -> bool
isOdd : int -> bool
isMultipleOfThree : int -> bool
endsWithZero : int -> bool
But you can use it differently. If you don't want to (or need to) create a new function, you can use it as follows:
isMultipleOf 10 150
This would return true, as 150 is multiple of 10. This is exactly the same as create the function endsWithZero and then applying it to the value 150.
Actually, function application is left associative, which means that the line above should be interpreted as:
(isMultipleOf 10) 150
That is, you put the parenthesis around the leftmost function application.
Now, if you can understand all this, your example (which is the canonical CreateAdder) should be trivial!
Sometime ago someone asked this question which deals with exactly the same concept, but in Javascript. In my answer I give two canonical examples (CreateAdder, CreateMultiplier) inf Javascript, that are somewhat more explicit about returning functions.
I hope this helps.
The canonical example of this is probably an "adder creator" - a function which, given a number (e.g. 3) returns another function which takes an integer and adds the first number to it.
So, for example, in pseudo-code
x = CreateAdder(3)
x(5) // returns 8
x(10) // returns 13
CreateAdder(20)(30) // returns 50
I'm not quite comfortable enough in F# to try to write it without checking it, but the C# would be something like:
public static Func<int, int> CreateAdder(int amountToAdd)
{
return x => x + amountToAdd;
}
Does that help?
EDIT: As Bruno noted, the example you've given in your question is exactly the example I've given C# code for, so the above pseudocode would become:
let x = f1 3
x 5 // Result: 8
x 10 // Result: 13
f1 20 30 // Result: 50
It's a function that takes an integer and returns a function that takes an integer and returns an integer.
This is functionally equivalent to a function that takes two integers and returns an integer. This way of treating functions that take multiple parameters is common in functional languages and makes it easy to partially apply a function on a value.
For example, assume there's an add function that takes two integers and adds them together:
let add x y = x + y
You have a list and you want to add 10 to each item. You'd partially apply add function to the value 10. It would bind one of the parameters to 10 and leaves the other argument unbound.
let list = [1;2;3;4]
let listPlusTen = List.map (add 10)
This trick makes composing functions very easy and makes them very reusable. As you can see, you don't need to write another function that adds 10 to the list items to pass it to map. You have just reused the add function.
You usually interpret this as a function that takes two integers and returns an integer.
You should read about currying.
a function taking an integer, which returns a function which takes an integer and returns an integer
The last part of that:
a function which takes an integer and returns an integer
should be rather simple, C# example:
public int Test(int takesAnInteger) { return 0; }
So we're left with
a function taking an integer, which returns (a function like the one above)
C# again:
public int Test(int takesAnInteger) { return 0; }
public int Test2(int takesAnInteger) { return 1; }
public Func<int,int> Test(int takesAnInteger) {
if(takesAnInteger == 0) {
return Test;
} else {
return Test2;
}
}
You may want to read
F# function types: fun with tuples and currying
In F# (and many other functional languages), there's a concept called curried functions. This is what you're seeing. Essentially, every function takes one argument and returns one value.
This seems a bit confusing at first, because you can write let add x y = x + y and it appears to add two arguments. But actually, the original add function only takes the argument x. When you apply it, it returns a function that takes one argument (y) and has the x value already filled in. When you then apply that function, it returns the desired integer.
This is shown in the type signature. Think of the arrow in a type signature as meaning "takes the thing on my left side and returns the thing on my right side". In the type int -> int -> int, this means that it takes an argument of type int — an integer — and returns a function of type int -> int — a function that takes an integer and returns an integer. You'll notice that this precisely matches the description of how curried functions work above.
Example:
let f b a = pown a b //f a b = a^b
is a function that takes an int (the exponent) and returns a function that raises its argument to that exponent, like
let sqr = f 2
or
let tothepowerofthree = f 3
so
sqr 5 = 25
tothepowerofthree 3 = 27
The concept is called Higher Order Function and quite common to functional programming.
Functions themselves are just another type of data. Hence you can write functions that return other functions. Of course you can still have a function that takes an int as parameter and returns something else. Combine the two and consider the following example (in python):
def mult_by(a):
def _mult_by(x):
return x*a
return mult_by
mult_by_3 = mult_by(3)
print mylt_by_3(3)
9
(sorry for using python, but i don't know f#)
There are already lots of answers here, but I'd like to offer another take. Sometimes explaining the same thing in lots of different ways helps you to 'grok' it.
I like to think of functions as "you give me something, and I'll give you something else back"
So a Func<int, string> says "you give me an int, and I'll give you a string".
I also find it easier to think in terms of 'later' : "When you give me an int, I'll give you a string". This is especially important when you see things like myfunc = x => y => x + y ("When you give curried an x, you get back something which when you give it a y will return x + y").
(By the way, I'm assuming you're familiar with C# here)
So we could express your int -> int -> int example as Func<int, Func<int, int>>.
Another way that I look at int -> int -> int is that you peel away each element from the left by providing an argument of the appropriate type. And when you have no more ->'s, you're out of 'laters' and you get a value.
(Just for fun), you can transform a function which takes all it's arguments in one go into one which takes them 'progressively' (the official term for applying them progressively is 'partial application'), this is called 'currying':
static void Main()
{
//define a simple add function
Func<int, int, int> add = (a, b) => a + b;
//curry so we can apply one parameter at a time
var curried = Curry(add);
//'build' an incrementer out of our add function
var inc = curried(1); // (var inc = Curry(add)(1) works here too)
Console.WriteLine(inc(5)); // returns 6
Console.ReadKey();
}
static Func<T, Func<T, T>> Curry<T>(Func<T, T, T> f)
{
return a => b => f(a, b);
}
Here is my 2 c. By default F# functions enable partial application or currying. This means when you define this:
let adder a b = a + b;;
You are defining a function that takes and integer and returns a function that takes an integer and returns an integer or int -> int -> int. Currying then allows you partiallly apply a function to create another function:
let twoadder = adder 2;;
//val it: int -> int
The above code predifined a to 2, so that whenever you call twoadder 3 it will simply add two to the argument.
The syntax where the function parameters are separated by space is equivalent to this lambda syntax:
let adder = fun a -> fun b -> a + b;;
Which is a more readable way to figure out that the two functions are actually chained.