program read_hex

implicit none
character(len=80) :: line="3 5  F6F  1d 0 0"
character(len=80) :: buff, hex_str, fm
integer, dimension(3) :: hex
integer :: str_len, i, first, last

buff = adjustl(line)
str_len = len_trim(buff)

!! remove the extra space in the string
do i=1,str_len-1
   if (buff(i:i) == ' ' .and. buff(i+1:i+1) == ' ') buff(i:) = buff(i+1:)
end do

!! extract the hexadecimal constant
first = 1
do i=1,3
   first = index(buff(first:), ' ') + first - 1
   last = index(buff(first+1:), ' ') + first
   hex_str = buff(first+1:last-1)
   write(fm, '(a,i0,a)') '(Z', len_trim(hex_str), ')'
   read(hex_str, fm) hex(i)
   first = last
end do

!! hexadecimal to decimal constrant, and output...
write(*,*) hex

end program read_hex

512396楼: Originally posted by snoopyzhao at 2011-07-21 09:58:42
看这个意思，是从6个16进制的数据中，读取其中的三个，然后返回3个对应的十进制数，是吧？

5楼: Originally posted by david830wu at 2011-07-18 13:08:23
嗯，我这个问题已经解决了：
Buffer = "3  2a  5  f0  0  0"  ! 就是说2a, 5, f0为三角网格的三个节点的索引，就是所要的数据，程序如下：
        !------------------------------------------------------- ...

11楼: Originally posted by snoopyzhao at 2011-07-21 11:45:11
给一个我写的 Fortran 90 版本的程序，没有加前面的准备字符串部分，呵呵，希望不难懂理解……


program read_hex

implicit none
character(len=80) :: line="3 5  F6F  1d 0 0"
character(len=80) :: buf ...