Using a deferred-length character string to read user input

Question

I would like to use deferred-length character strings in a "simple" manner to read user input. The reason that I want to do this is that I do not want to have to declare the size of a character string before knowing how large the user input will be. I know that there are "complicated" ways to do this. For example, the iso_varying_string module can be used: https://www.fortran.com/iso_varying_string.f95. Also, there is a solution here: Fortran Character Input at Undefined Length. However, I was hoping for something as simple, or almost as simple, as the following:

program main

  character(len = :), allocatable :: my_string
  read(*, '(a)') my_string
  write(*,'(a)') my_string
  print *, allocated(my_string), len(my_string)

end program

When I run this program, the output is:

./a.out
here is the user input

F       32765

Notice that there is no output from write(*,'(a)') my_string. Why?

Also, my_string has not been allocated. Why?

Why isn't this a simple feature of Fortran? Do other languages have this simple feature? Am I lacking some basic understanding about this issue in general?

Answer 1

vincentjs's answer isn't quite right.

Modern (2003+) Fortran does allow automatic allocation and re-allocation of strings on assignment, so a sequence of statements such as this

character(len=:), allocatable :: string
...
string = 'Hello'
write(*,*)
string = 'my friend'
write(*,*)
string = 'Hello '//string
write(*,*)

is correct and will work as expected and write out 3 strings of different lengths. At least one compiler in widespread use, the Intel Fortran compiler, does not engage 2003 semantics by default so may raise an error on trying to compile this. Refer to the documentation for the setting to use Fortran 2003.

However, this feature is not available when reading a string so you have to resort to the tried and tested (aka old-fashioned if you prefer) approach of declaring a buffer of sufficient size for any input and of then assigning the allocatable variable. Like this:

character(len=long) :: buffer
character(len=:), allocatable :: string
...
read(*,*) buffer
string = trim(buffer)

No, I don't know why the language standard forbids automatic allocation on read, just that it does.