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Re if / then statement

 
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Zach



Joined: 13 Mar 2023
Posts: 85
Location: Groningen, Netherlands

PostPosted: Sat May 13, 2023 1:32 pm    Post subject: Re if / then statement Reply with quote

How do I check whether a character is a space, in an if / then statement?

Last edited by Zach on Sat May 13, 2023 6:03 pm; edited 4 times in total
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JohnCampbell



Joined: 16 Feb 2006
Posts: 2532
Location: Sydney

PostPosted: Sat May 13, 2023 2:13 pm    Post subject: Reply with quote

The following changes give no compiler errors, although I am not sure if the code does what you described.

There is more work to read other lines ?

Code:
program test
character(200) :: stringin
character,dimension(200) :: strgout
integer :: single_character, i

stringin = 'the kat kissed the dog and ate his banana. The banana was too big for its mouth so it had to sneeze. '
do i = 1, len(stringin)
  single_character = iachar(stringin(i:i))
  strgout(i) = achar(single_character)
  strgout(i) = stringin(i:i)        ! this should be the same as the previous 2 lines
  if (i > 40) then
    if (strgout(i) == achar(32)) print *,'\n'    ! are these prints what you want ?
  end if
end do

!  what do you want to do with strgout ?

end program


note that both "character(200) :: stringin" and "character,dimension(200) :: strgout" can be used to achieve a similar outcome, with appropriate Fortran syntax, so converting from one to the other can be avoided.
"stringin" probably has easier syntax when being used.
You should review intrinsic routines LEN, LEN_TRIM and TRIM when using stringin.
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Zach



Joined: 13 Mar 2023
Posts: 85
Location: Groningen, Netherlands

PostPosted: Sat May 13, 2023 9:01 pm    Post subject: Reply with quote

The objective is to list strings 40 characters wide
The skip back up isn't working. Please help.

program test
character(200) :: strIn
integer::single_character

strIn= 'the kat kissed the dog and ate his banana. The banana was too big for its mouth so it had to sneeze. '

do i = 1, len(strIn), 1
single_character = ichar(strIn(i:i))
write(*,*) achar(single_character)
if (i < 40) print *, achar(43)!skip back up

end do
end program
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JohnCampbell



Joined: 16 Feb 2006
Posts: 2532
Location: Sydney

PostPosted: Wed May 17, 2023 3:27 am    Post subject: Reply with quote

This is not the final position, but a start ?

Code:
program test
character(200) :: strIn
integer :: i,k

strIn= 'the kat kissed the dog and ate his banana. The banana was too big for its mouth so it had to sneeze. '

do i = 1, len(strIn), 40
 k = min(len(strIn),i+39)
 write(*,*) strIn( i:k )
end do
end program


You could write code to modify "k" for the end of a "word"
Code:

program test
character(200) :: strIn
integer :: i,k,n

strIn= 'the kat kissed the dog and ate his banana. The banana was too big for its mouth so it had to sneeze. '

 n = len_trim (strIn)
 i = 1
 do
   k = min (n,i+39)
   k = i + INDEX ( strIn(i:k+1), ' ', .true. ) - 1
   if ( k > i ) write(*,*) strIn( i:k )
   i = k+1
   if ( i > n ) exit
end do
end program


There may be more tests for loop to end or if there are very long words !
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Kenneth_Smith



Joined: 18 May 2012
Posts: 696
Location: Hamilton, Lanarkshire, Scotland.

PostPosted: Wed May 17, 2023 10:38 am    Post subject: Reply with quote

Does this help? Space_loc(N) holds the location of the Nth occurrence of ichar(' ')

Code:
program test
implicit none
character(200) :: stringin
integer, dimension(200) :: space
integer, dimension(200) :: space_loc
integer i, j, counter, total

stringin = ' the kat kissed the dog and ate his banana. The banana was too big for its mouth so it had to sneeze. '

do i = 1, len(stringin), 1
  if (ichar(stringin(i:i)) .eq. ichar(' ')) then
    space(i) = 1   ! True
  else
    space(i) = 0   ! False
  end if
end do
 
total     = sum(space)
counter   = 1
space_loc = 0
j = 1
do i = 1, len_trim(stringin), 1
  if (space(i) .eq. 1) then
    space_loc(j) = counter
    j = j + 1
  end if
  counter = counter + 1
end do

! space_loc(N) gives the location of the Nth occurrence of ichar(' ')

  do i = 1, total , 1
    if (space_loc(i) .gt. 0) print*, "ichar(' ')", i,  " at location ", space_loc(i)
  end do
end
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Zach



Joined: 13 Mar 2023
Posts: 85
Location: Groningen, Netherlands

PostPosted: Wed May 17, 2023 1:44 pm    Post subject: Reply with quote

Re the responses to my question! I am very grateful. Thank you! Patrick.
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Kenneth_Smith



Joined: 18 May 2012
Posts: 696
Location: Hamilton, Lanarkshire, Scotland.

PostPosted: Wed May 17, 2023 3:02 pm    Post subject: Reply with quote

Generalising my previous post:
Code:
module string_mod
implicit none
integer, protected, allocatable :: location(:)
contains
  subroutine locatechar(instring,searchchar,number)
  character(len=*), intent(in) :: instring
  character(len=1), intent(in) :: searchchar
  integer, intent(out)         :: number
  integer :: a(len(instring))
  integer i, j, counter
    a      = 0
    number = 0
    do i = 1, len(instring), 1
      if (ichar(instring(i:i)) .eq. ichar(searchchar))then
        a(i)   = 1
        number = number + 1
      end if
    end do   
    if (.not. allocated(location)) then
      allocate(location(len(instring)))
    else if (allocated(location) .and. size(location) .ne. len(instring)) then
      deallocate(location)
      allocate(location(len(instring)))
    end if
    location  = 0
    counter   = 1
    j = 1
    do i = 1, len_trim(instring), 1
      if (a(i) .eq. 1) then
        location(j) = counter
        j = j + 1
      end if
      counter = counter + 1
    end do
  end subroutine locatechar
end module string_mod

program test
use string_mod
implicit none
character(200) :: stringin
character(50)  :: string2
integer  total, i
stringin = ' the kat kissed the dog and ate his banana. The banana was too big for its mouth so it had to sneeze. '
string2 = 'Casper is a fiesty terrior'
call locatechar(stringin,'a',total)
print'(/a)',      repeat('1234567890',10)
print'(a)',       repeat('         |',10)
print'(a)',       trim(stringin)
print'(a,1x,i3)', "Occurrences of 'a' ", total
do i = 1, total , 1
    print*, "'k'", i,  " at location ", location(i)
end do
call locatechar(string2,'r', total)
print'(/a)',      repeat('1234567890',10)
print'(a)',       repeat('         |',10)
print'(a)',       trim(string2)
print'(a,1x,i3)', "Occurrences of 'r' ", total
  do i = 1, total , 1
 print*, "'r'", i,  " at location ", location(i)
end do
end
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