Week 8

Today’s program

New stuff:– Recursion in F– Procedures as arguments– User-defined data types

Quiz #3

What is recursion?

Recursion in programming is a technique for defining a problem in terms of one or more smaller versions of the same problem.

The solution to the problem is built on the result(s) from the smaller version(s).

What is recursion?

Best known example:the factorial

n! = n (n-1)!0! = 1

That’s all you need to calculate n!

Recursive procedures

recursive function function_name(…) result(result)

Let’s use it to calculate n! Structure plan

• n = 0factorial_n = 1

• n > 0factorial_n = n * factorial(n-1)

• n < 0 Error! Return factorial_n = 0

Solution 1 recursive function factorial(n) result(factorial_n)

! Dummy argument and result variable integer, intent(in) :: n real :: factorial_n ! Determine whether further recursion is required select case(n) case(0) ! Recursion has reached the end factorial_n = 1.0 case(1:) ! More recursive calculation(s) required factorial_n = n*factorial(n-1) case default ! n is negative - return zero as an error indicator factorial_n = 0.0end function factorial

Solution 2 recursive subroutine factorial(n, factorial_n)

! Dummy arguments integer, intent(in) :: n real, intent(out) :: factorial_n ! Determine whether further recursion is required select case(n) case(0) ! Recursion has reached the end factorial_n = 1.0 case(1:) ! Recursive call(s) required ! to obtain (n-1)! call factorial(n-1, factorial_n) case default ! n is negative - return zero as an error indicator factorial_n = 0.0end subroutine factorial

Procedures as arguments

Up to now, procedures had as dummy arguments

real, integer, character, logical

andarrays with these types

But, what about a procedure as a dummy argument?

How to declare a procedure?

We have to provide information concerning the interface of the procedure

The interface block:interface

interface_bodyend interface

How to declare a procedure?

Example:

interface function dummy_fun(a, b) result(r) real, intent(in) :: a, b

real :: r end function dummy_funend interface

How to declare a procedure?

Example:

interface subroutine one_arg(x) real, intent(inout) :: x end subroutine one_arg recursive subroutine two_args(x, y) real, intent(inout) :: x, y end subroutine two_argsend interface

Couple of details...

Dummy procedure arguments in a function must be functions

All actual procedure arguments must be module procedures

It is not permissible for an intrinsic procedure to be an actual argument

Derived data types

You can create your own data types! They can be derived from:

– Intrinsic data types:integer, real, character,

logical

– Previously defined data types

Derived data types

type, public :: new_type component_definition . . .end type new_type

Example

type, public :: person

character(len=12) :: first_name

character(len=1) :: middle_initial

character(len=12) :: last_name

integer :: age

character(len=1) :: sex ! M or F

character(len=5) :: tax_number

end type person

Example

Now you can declare variable of type person:

type(person) :: ali, mukaddes, veli

Derived types vs. arrays

1 2 3 4 5 6

A

Be careful...

Declarations need access to the type definition

Place definitions in a module

Putting data in a derived type variable

ali = person("Ali","M","Aktepe",56,"M","45645")

mukaddes =

person("Mukaddes"," ","Has",18,"F","12345")

veli = person("Veli","M","Karatepe",65,"M","34567")

This is a structure constructor

Refering to components

variable%component Example:

ali%last_namemukaddes%ageveli%first_name

Using derived types to define new derived types type, public :: employee type(person) :: employee character(len=20) department real :: salaryend type employee

type(person) :: saadet saadet%employee%age = 34

Arrays of derived types type(person), dimension(1:12) ::

bil102class

Then:

bil102class(2)%sex = "M"

You can do operations on components:

bil102class(3)%age+bil102class(4)%age But no such operation is permissible:

bil102class(3) - bil102class(4)

Example:Let's do some geometry! Two derived types:

– point– line

Two functions:– distinct_points(p1,p2) result(distinct)– line_from_points(p1,p2) result(join_line)

Assignment due on April 22

Problem 8.3 from Ellis & Philips Make sure to fully document your

programs!