Say I want to parse the following string:
"Compute the sum of (integer) and (integer)" in Fortran 90, of which I have no way of telling how large the integer will be. It can be 3, just as well as 300,000.
As far as I can tell the FORMAT statement does not leave room for inferring the size of an integer at run-time. Choose a too large size, say i5, for a number as small as 3 however and the program crashes.
How would I best go about this?
If the location of integers in the string is known at compile time (e.g. 5th and 7th words), we can get the integers directly using list-directed read:
character(256) :: line, words( 50 )
integer :: i1, i2
line = "Compute the sum of 3 and 30000, then we get..."
read( line, * ) words( 1 : 7 ) !! get the first seven words from a line
read( words( 5 ), * ) i1 !! convert the 5th and 7th words to integer
read( words( 7 ), * ) i2 !! so that i1 = 3, i2 = 30000
But if the location of integers is not known (e.g. when obtained from user input), things are probably more complicated... I have written some subroutines for this, so if it may seem useful please try it :)
module strmod
contains
subroutine split ( line, words, nw )
implicit none
character(*), intent(in) :: line
character(*), intent(out) :: words(:)
integer, intent(out) :: nw
character(len(words)) :: buf( size(words) )
integer :: k, ios
nw = 0 ; words(:) = ""
do k = 1, size(words)
read( line, *, iostat=ios ) buf( 1 : k )
if ( ios /= 0 ) exit
nw = k
words( 1 : nw ) = buf( 1 : nw )
enddo
endsubroutine
subroutine words_to_ints ( words, ints, ni )
implicit none
character(*), intent(in) :: words(:)
integer, intent(out) :: ints(:)
integer, intent(out) :: ni
integer :: k, val, ios
ni = 0 ; ints(:) = 0
do k = 1, size(words)
read( words( k ), *, iostat=ios ) val
if ( ios /= 0 ) cycle
ni = ni + 1
if ( ni > size(ints) ) stop "size(ints) too small"
ints( ni ) = val
enddo
endsubroutine
endmodule
program main
use strmod
implicit none
character(80) :: line, words( 50 ) !! works also with size 5 or 7 etc
integer :: ints( 50 ), nw, ni, k
line = "Compute the sum of 3 and 30000, then we get 300003 (no!!)"
!... Note: spaces and commas serve as delimiters. Better to avoid "/".
call split ( line, words, nw )
call words_to_ints ( words, ints, ni )
print *, "Word counts:", nw
do k = 1, nw
print *, trim( words( k ) )
enddo
print *, "Int counts:", ni
print *, ints( 1 : ni )
end
Results:
Word counts: 12
Compute
the
sum
of
3
and
30000
then
we
get
300003
(no!!)
Int counts: 3
3 30000 300003
