High Performance Mark's comment tells you about why the compiler complains: implicit typing.
The result of the function concat is implicitly typed because you haven't declared its type otherwise. Although x // 'plus stuff' is the correct way to concatenate character variables, you're attempting to assign that new character object to a (implictly) real function result.
Which leads to the question: "just how do I declare the function result to be a character?". Answer: much as you would any other character variable:
character(len=length) concat
[note that I use character(len=...) rather than character*.... I'll come on to exactly why later, but I'll also point out that the form character*4 is obsolete according to current Fortran, and may eventually be deleted entirely.]
The tricky part is: what is the length it should be declared as?
When declaring the length of a character function result which we don't know ahead of time there are two1 approaches:
- an automatic character object;
- a deferred length character object.
In the case of this function, we know that the length of the result is 10 longer than the input. We can declare
character(len=LEN(x)+10) concat
To do this we cannot use the form character*(LEN(x)+10).
In a more general case, deferred length:
character(len=:), allocatable :: concat ! Deferred length, will be defined on allocation
where later
concat = x//'plus stuff' ! Using automatic allocation on intrinsic assignment
Using these forms adds the requirement that the function concat has an explicit interface in the main program. You'll find much about that in other questions and resources. Providing an explicit interface will also remove the problem that, in the main program, concat also implicitly has a real result.
To stress:
program
implicit none
character(len=[something]) concat
print *, concat('hell')
end program
will not work for concat having result of the "length unknown at compile time" forms. Ideally the function will be an internal one, or one accessed from a module.
1 There is a third: assumed length function result. Anyone who wants to know about this could read this separate question. Everyone else should pretend this doesn't exist. Just like the writers of the Fortran standard.
concatand Fortran has gone right ahead and used good old implicit typing to determine that it is arealfunction whereas you have tried to assign a string to its return variable. Any reasonable Fortran reference, and a host of SO qs and as, will explain the mechanics of implicit typing and that it is evil. Beyond that you've not ensured theprogramhas access to the interface of the function, so the program also thinks thatconcatis a real-valued thing. Do more research into the use ofcontainsand ofmodulein Fortran.implicit none. It will make your life so much easier.