recursion using for do loop (pascal)

Question

I'm trying to use the concept of recursion but using for do loop. However my program cannot do it. For example if I want the output for 4! the answer should be 24 but my output is 12. Can somebody please help me?

 program pastYear;

 var
   n,i:integer;


 function  calculateFactorial ( A:integer):real;

 begin
   if A=0 then
     calculateFactorial := 1.0
   else
     for i:= A downto 1 do
     begin
       j:= A-1;
       calculateFactorial:= A*j;
     end;
   end;

   begin
     writeln( ' Please enter a number ');
     readln ( n);
     writeln ( calculateFactorial(n):2:2);
     readln;
   end.

j isn't declared anywhere. Also, A never changes in your loop, therefore j (from j := A-1 never changes, and, therefore, A*j never changes. Your factorial program returns A*(A-1). — lurker
Recursion would be A*calculateFactorial(A-1), no loop necessary. — D. Ben Knoble
Guys thanks for helping me out. I finally got the answer I've been looking for. About the loop I know that we can do recursion without it but the question from my lecturer want us to use the loop so that's why. — exo_1

Wosi Wosi · Accepted Answer · 2015-10-22T09:19:18

There are several problems in your code.

First of all it doesn't compile because you are accessing the undefined variable j.
Calculating the factorial using a loop is the iterative way of doing it. You are looking for the recursive way.

What is a recursion? A recursive function calls itself. So in your case calculateFactorial needs a call to itself.

How is the factorial function declared?

In words:

The factorial of n is declared as

1 when n equals 0

the factorial of n-1 multiplied with n when n is greater than 0

So you see the definition of the factorial function is already recursive since it's referring to itself when n is greater than 0.

This can be adopted to Pascal code:

function Factorial(n: integer): integer;
begin
  if n = 0 then
    Result := 1
  else if n > 0 then
    Result := Factorial(n - 1) * n; 
end;

No we can do a few optimizations:

The factorial function doesn't work with negative numbers. So we change the datatype from integer (which can represent negative numbers) to longword (which can represent only positive numbers).
The largest value that a longword can store is 4294967295 which is twice as big as a longint can store.
Now as we don't need to care about negative numbers we can reduce one if statement.

The result looks like this:

function Factorial(n: longword): longword;
begin
  if n = 0 then
    Result := 1
  else 
    Result := Factorial(n - 1) * n; 
end;

recursion using for do loop (pascal)

1 Answers