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Functions (F#)

Functions are the fundamental unit of program execution in any programming language. As in other languages, an F# function has a name, can have parameters and take arguments, and has a body. F# also supports functional programming constructs such as treating functions as values, using unnamed functions in expressions, composition of functions to form new functions, curried functions, and the implicit definition of functions by way of the partial application of function arguments.

You define functions by using the let keyword, or, if the function is recursive, the let rec keyword combination.

// Non-recursive function definition.
let [inline] function-name parameter-list [ : return-type ] = function-body
// Recursive function definition.
let rec function-name parameter-list = recursive-function-body

The function-name is an identifier that represents the function. The parameter-list consists of successive parameters that are separated by spaces. You can specify an explicit type for each parameter, as described in the Parameters section. If you do not specify a specific argument type, the compiler attempts to infer the type from the function body. The function-body consists of an expression. The expression that makes up the function body is typically a compound expression consisting of a number of expressions that culminate in a final expression that is the return value. The return-type is a colon followed by a type and is optional. If you do not specify the type of the return value explicitly, the compiler determines the return type from the final expression.

A simple function definition resembles the following:

let f x = x + 1

In the previous example, the function name is f, the argument is x, which has type int, the function body is x + 1, and the return value is of type int.

The inline specifier is a hint to the compiler that the function is small and that the code for the function can be integrated into the body of the caller.

At any level of scope other than module scope, it is not an error to reuse a value or function name. If you reuse a name, the name declared later shadows the name declared earlier. However, at the top level scope in a module, names must be unique. For example, the following code produces an error when it appears at module scope, but not when it appears inside a function:

let list1 = [ 1; 2; 3]
// Error: duplicate definition. 
let list1 = []  
let function1 =
   let list1 = [1; 2; 3]
   let list1 = []
   list1

But the following code is acceptable at any level of scope:

let list1 = [ 1; 2; 3]
let sumPlus x =
// OK: inner list1 hides the outer list1. 
   let list1 = [1; 5; 10]  
   x + List.sum list1

Names of parameters are listed after the function name. You can specify a type for a parameter, as shown in the following example:

let f (x : int) = x + 1

If you specify a type, it follows the name of the parameter and is separated from the name by a colon. If you omit the type for the parameter, the parameter type is inferred by the compiler. For example, in the following function definition, the argument x is inferred to be of type int because 1 is of type int.

let f x = x + 1

However, the compiler will attempt to make the function as generic as possible. For example, note the following code:

let f x = (x, x)

The function creates a tuple from one argument of any type. Because the type is not specified, the function can be used with any argument type. For more information, see Automatic Generalization (F#).

A function body can contain definitions of local variables and functions. Such variables and functions are in scope in the body of the current function but not outside it. When you have the lightweight syntax option enabled, you must use indentation to indicate that a definition is in a function body, as shown in the following example:

let cylinderVolume radius length =
    // Define a local value pi. 
    let pi = 3.14159
    length * pi * radius * radius

For more information, see Code Formatting Guidelines (F#) and Verbose Syntax (F#).

The compiler uses the final expression in a function body to determine the return value and type. The compiler might infer the type of the final expression from previous expressions. In the function cylinderVolume, shown in the previous section, the type of pi is determined from the type of the literal 3.14159 to be float. The compiler uses the type of pi to determine the type of the expression h * pi * r * r to be float. Therefore, the overall return type of the function is float.

To specify the return value explicitly, write the code as follows:


let cylinderVolume radius length : float =
   // Define a local value pi. 
   let pi = 3.14159
   length * pi * radius * radius

As the code is written above, the compiler applies float to the entire function; if you mean to apply it to the parameter types as well, use the following code:

let cylinderVolume (radius : float) (length : float) : float

You call functions by specifying the function name followed by a space and then any arguments separated by spaces. For example, to call the function cylinderVolume and assign the result to the value vol, you write the following code:

let vol = cylinderVolume 2.0 3.0

If you supply fewer than the specified number of arguments, you create a new function that expects the remaining arguments. This method of handling arguments is referred to as currying and is a characteristic of functional programming languages like F#. For example, suppose you are working with two sizes of pipe: one has a radius of 2.0 and the other has a radius of 3.0. You could create functions that determine the volume of pipe as follows:

let smallPipeRadius = 2.0
let bigPipeRadius = 3.0

// These define functions that take the length as a remaining 
// argument: 

let smallPipeVolume = cylinderVolume smallPipeRadius
let bigPipeVolume = cylinderVolume bigPipeRadius

You would then supply the additional argument as needed for various lengths of pipe of the two different sizes:

let length1 = 30.0
let length2 = 40.0
let smallPipeVol1 = smallPipeVolume length1
let smallPipeVol2 = smallPipeVolume length2
let bigPipeVol1 = bigPipeVolume length1
let bigPipeVol2 = bigPipeVolume length2

Recursive functions are functions that call themselves. They require that you specify the rec keyword following the let keyword. Invoke the recursive function from within the body of the function just as you would invoke any function call. The following recursive function computes the nth Fibonacci number. The Fibonacci number sequence has been known since antiquity and is a sequence in which each successive number is the sum of the previous two numbers in the sequence.

let rec fib n = if n < 2 then 1 else fib (n - 1) + fib (n - 2)

Some recursive functions might overflow the program stack or perform inefficiently if you do not write them with care and with awareness of special techniques, such as the use of accumulators and continuations.

In F#, all functions are considered values; in fact, they are known as function values. Because functions are values, they can be used as arguments to other functions or in other contexts where values are used. Following is an example of a function that takes a function value as an argument:

let apply1 (transform : int -> int ) y = transform y

You specify the type of a function value by using the -> token. On the left side of this token is the type of the argument, and on the right side is the return value. In the previous example, apply1 is a function that takes a function transform as an argument, where transform is a function that takes an integer and returns another integer. The following code shows how to use apply1:

let increment x = x + 1

let result1 = apply1 increment 100

The value of result will be 101 after the previous code runs.

Multiple arguments are separated by successive -> tokens, as shown in the following example:

let apply2 ( f: int -> int -> int) x y = f x y

let mul x y = x * y

let result2 = apply2 mul 10 20

The result is 200.

A lambda expression is an unnamed function. In the previous examples, instead of defining named functions increment and mul, you could use lambda expressions as follows:

let result3 = apply1 (fun x -> x + 1) 100

let result4 = apply2 (fun x y -> x * y ) 10 20

You define lambda expressions by using the fun keyword. A lambda expression resembles a function definition, except that instead of the = token, the -> token is used to separate the argument list from the function body. As in a regular function definition, the argument types can be inferred or specified explicitly, and the return type of the lambda expression is inferred from the type of the last expression in the body. For more information, see Lambda Expressions: The fun Keyword (F#).

Functions in F# can be composed from other functions. The composition of two functions function1 and function2 is another function that represents the application of function1 followed the application of function2:

let function1 x = x + 1
let function2 x = x * 2
let h = function1 >> function2
let result5 = h 100

The result is 202.

Pipelining enables function calls to be chained together as successive operations. Pipelining works as follows:

let result = 100 |> function1 |> function2

The result is again 202.

The composition operators take two functions and return a function; by contrast, the pipeline operators take a function and an argument and return a value. The following code example shows the difference between the pipeline and composition operators by showing the differences in the function signatures and usage.

// Function composition and pipeline operators compared.

let addOne x = x + 1
let timesTwo x = 2 * x

// Composition operator
// ( >> ) : ('T1 -> 'T2) -> ('T2 -> 'T3) -> 'T1 -> 'T3
let Compose2 = addOne >> timesTwo

// Backward composition operator
// ( << ) : ('T2 -> 'T3) -> ('T1 -> 'T2) -> 'T1 -> 'T3
let Compose1 = addOne << timesTwo

// Result is 5
let result1 = Compose1 2

// Result is 6
let result2 = Compose2 2

// Pipelining
// Pipeline operator
// ( <| ) : ('T -> 'U) -> 'T -> 'U
let Pipeline1 x = addOne <| timesTwo x

// Backward pipeline operator
// ( |> ) : 'T1 -> ('T1 -> 'U) -> 'U
let Pipeline2 x = addOne x |> timesTwo

// Result is 5
let result3 = Pipeline1 2

// Result is 6
let result4 = Pipeline2 2

You can overload methods of a type but not functions. For more information, see Methods (F#).

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