Function result has no implicit type

Question

Below is a sample code that addresses the problem I am having. The error message I am getting is

Function result \'sample\' at (1) has no IMPLICIT type.<

Answer 1

Procedures in Fortran come in two types: functions and subroutines. This question is about functions, so I'll consider just those.

What was missing in the first revision, giving the error about the implicit type of the function result¹, was the result type.

Adding real function ... or complex function ..., etc., resolves that problem by explicitly giving the type of the function result. The linked documentation gives other ways of doing that.

The function's result is used when the function is referenced. When we have a reference like

func0 = Sample(func)

in the main program, the function Sample is invoked and the function result is defined in its execution. At the end of the function's execution its result is placed in the expression of the reference.

So, if you declare

real function Sample(func)

or

complex function Sample(func)

what you are saying is that the function result is either a real or complex entity. And when the function is evaluated, whatever value Sample had at the end is used in the expression (here assignment).

As a consequence of the function result being returned through Sample (in this case) we need to define its value. The important thing to note for the question, then, is that LocalF is a variable local to the function. If you mean it to be the result of the function you need to use the function result.

You have a number of options:

function Sample(func)
  <type>, <attributes> :: sample  ! Instead of LocalF
  ...                  :: func
end function

or

function Sample(func) result(LocalF)
  <type>, <attributes> :: LocalF
  ...                  :: func
end function

You can even have

<type> function Sample(func)
  <attribute statements for Sample>
  ... func
end function

but I really suggest you avoid that last one.

---

¹ Note the error here is about type for the function result; in the linked question simply about the function when referenced.

Answer 2

In Fortran every function has a result. If you like you can think of the result as a value returned by the function. Like every other value in a Fortran program a function result has a type, and a kind and a rank too.

By default the function result has the same name as the function itself, and its declaration is prepended to the function declaration. For example, here

integer function add(m,n)
    integer, intent(in) :: a,b
    add = a+b
end function

the function is called add and you can see (a) that the result is of type integer (and of default kind and scalar) and (b) that the result is formed by adding the two arguments together.

For functions returning arrays this syntax is not available, so you couldn't write something like

 integer(1:4) add_vec(m1,m2)

In such cases you have to explicitly define the name (and later type and kind) of the result variable. Sticking with the simple example, something like

   function add(m,n) result(addvec)
        integer, intent(in) :: a(4),b(4)
        integer, dimension(4) :: addvec
        ....
    end function

Notice that you don't define the intent of the result.

In OP's case sample is, I think, intended to return a rank-2 array of complex values. I think OP needs to replace

function Sample(func)   !this is line (1)

with

function Sample(func)  result(LocalF)

and see how that goes. Here, if it is not evident already, you learn that the result name doesn't have to be the same as the name of the function.

Furthermore ... Adding Real or Complex in front of function works, but I don't really get why.

It might work in the sense of compiling, but executing it will lead to tears. By telling the compiler that the function result is either a real or complex value you satisfy the syntactical requirements for a function definition. But without assigning a (real or complex as declared) value to the result variable (called Sample in OP's code) the function will, at best, return junk.

To be as clear as I can ... in OP's original code there were two serious mistakes:

1. The function (result) was not given an explicit type, which lead to the compiler message shown.
2. The function did not include setting the value of the result variable, i.e. the variable with the same name as the function (in the absence of the result clause).