Removing whitespace in string - fortran

Question

I have the following code:
program main
character (len=15) :: abc = "te st tex t"
print *, trim(abc)
end program main
Which outputs:
te st tex t
I excepted all the whitespace to be removed but it wasn't. How can I remove all the whitespace from the string?

Answer 1

Trim will remove spaces only at the edges, not in the middle (this is common behaviour on almost all languages/libraries). If you want to remove all spaces in the string, you will have to create your own function to do this, iterating through the string.
Ex.:
program Test
implicit none
! Variables
character(len=200) :: string
! Body of Test
string = 'Hello World 7 9'
print *, string
call StripSpaces (string)
print *, string
contains
subroutine StripSpaces(string)
character(len=*) :: string
integer :: stringLen
integer :: last, actual
stringLen = len (string)
last = 1
actual = 1
do while (actual < stringLen)
if (string(last:last) == ' ') then
actual = actual + 1
string(last:last) = string(actual:actual)
string(actual:actual) = ' '
else
last = last + 1
if (actual < last) &
actual = last
endif
end do
end subroutine
end program Test
This was tested on intel compiler, not on gfortran, but I think it will work.

Answer 2

I was able to do this using the variable string library described here ( http://schonfelder.co.uk/is1539-2-99.htm ). The source code link is found in the introduction section of the ISO document.
Here is the code
program Console1
use ISO_VARYING_STRING
implicit none
! Body of Console1
character(LEN=50) :: text = 'Hello World John Mary '
character(LEN=50) :: res
print *, trim(text)
! 'Hello World John Mary'
res = REPLACE(text,' ','', every=.TRUE.)
print *, trim(res)
! 'HelloWorldJohnMary'
end program Console1

Answer 3

Here's a dirty, shameful way to eliminate the spaces. This is only likely to work if a compiler lays out a length-15 string in the same order and space as it would a 15-element array of characters. While this is likely to be true, and in my recent experience is true, it is not guaranteed to be so by the standard. That aside, this approach may be good enough.
! declarations
CHARACTER (len=15) :: abc = "te st tex t"
CHARACTER, DIMENSION(LEN(abc)) :: abc_array
! or CHARACTER, DIMENSION(:), ALLOCATABLE :: abc_array if your compiler supports
! automatic allocation
! transfer the string into an array of characters
abc_array = TRANSFER(abc,abc_array)
! eliminate the spaces, and transfer back to the string
abc = TRANSFER(PACK(abc_array,abc_array/=' '),abc)
! now all the spaces are at the end of abc so the following statement writes the
! string with no spaces
WRITE(*,*) TRIM(abc)
Use this approach at your own risk.

Answer 4

For those averse to TRANSFER perhaps a nice little recursive function would appeal. As written this depends on Fortran 2003's ability to automatically allocate character scalars, but it shouldn't be too hard to modify if your compiler doesn't support this feature yet.
RECURSIVE FUNCTION stripper(string,ch) RESULT(stripped)
CHARACTER(len=*), INTENT(in) :: string
CHARACTER, INTENT(in) :: ch
CHARACTER(:), ALLOCATABLE :: stripped
IF (LEN(string)==1) THEN
IF (string==ch) THEN
stripped = ''
ELSE
stripped = string
END IF
ELSE
IF (string(1:1)==ch) THEN
stripped = stripper(string(2:),ch)
ELSE
stripped = string(1:1)//stripper(string(2:),ch)
END IF
END IF
END FUNCTION stripper

Answer 5

You can try this:
program test
!erase blank space in a string
!run over every character of the string and just take every non-blank in other variable.
implicit none
character (len=100) str1,str2
integer i
str2='' !in this variable will be save non-blank spaces
str1=' a b c de ' !Test string with blank spaces
write(*,*)len_trim(str1), str1
do i=1,len(str1)
if (str1(i:i).ne.' ')str2=trim(str2)//trim(str1(i:i))
end do
write(*,*)len_trim(str2), str2
end