Fortran 77 从输入中获取字符串直到 space 就像 java 中的 input.next()

Question

我是 Fortran 77 的新手。我需要像 Java 中的 input.next() 这样的代码。我想给出一个像 "hi how are you today" 这样的输入并一个一个地检查每个单词，所以要做到这一点，我需要一个一个地接受单词。最简单的方法是什么？我可以检查每个字符并将字符放在字符变量 space 之前，但它看起来很难。

Answer 1

虽然这可能是重复的，但这是我手头的东西（表明没有直接的内置例程，虽然乍一看可能很棘手，但并不是 "hard"以某种方式写它......）我认为还有基于 index() 函数的更有效的方法。如果有必要，写一个类似 "input_next()" 的例程来获取下一个单词 one-by-one [*].

可能会有用

program main
    implicit none
    character(100) line, words( 100 )
    integer n, i

    line = "hi how are you today"   ! input string
    words = ""                      ! words to be obtained

    call split_str( line, words, n )

    do i = 1, n
        print *, "i= ", i, "word= ", trim(words( i ))
    enddo
end

subroutine split_str( line, words, n )
    implicit none
    character(*), intent(in)  :: line
    character(*), intent(out) :: words(*)
    integer,      intent(out) :: n
    integer :: ios
    character(100) :: buf( 100 )  ! large buffer

    n = 0
    do
        n = n + 1
        read( line, *, iostat=ios ) buf( 1 : n )  ! use list-directed input
        if ( ios == 0 ) then
            words( 1 : n ) = buf( 1 : n )   ! if success, copy to the original array
        else
            n = n - 1
            exit       ! if all the words are obtained, finish
        endif
    enddo
end

结果：

 i=            1 word= hi
 i=            2 word= how
 i=            3 word= are
 i=            4 word= you
 i=            5 word= today

[*] 这是这种 getnextword() 的一种可能方法，它通过 list-directed 输入从输入字符串 (line) 中获取一个词，然后从字符串中删除该词下次通话。如果在 line 中找不到更多的单词，则 found 变为假。（请在网络或 SO 页面中搜索 "list-directed input" 以获取更多详细信息。）

program main
    implicit none
    character(100) line, word
    logical found

    line = "hi how are you today"
    do
        call getnextword( line, word, found )
        if ( .not. found ) exit
        print "(a,a7,2a)", "word= ", trim( word ), " : line= ", trim( line )
    enddo
end program

subroutine getnextword( line, word, found )
    implicit none
    character(*), intent(inout) :: line
    character(*), intent(out)   :: word
    logical,      intent(out)   :: found
    integer :: ios
    character(100) :: buf

    read( line, *, iostat=ios ) buf   ! try to read one word into a buffer via list-directed input

    if ( ios == 0 ) then       ! if success
        found = .true.
        word = trim( buf )      ! save the word
        line = adjustL( line )
        line = line( len_trim( word ) + 1 : )   ! and remove the word from the input line
    else
        found = .false.
        word = ""
    endif
end

结果：

word=      hi : line=  how are you today
word=     how : line=  are you today
word=     are : line=  you today
word=     you : line=  today
word=   today : line=

Fortran 77 从输入中获取字符串直到 space 就像 java 中的 input.next()

fortran 77 take string from input until space like input.next() in java

fortran

input

fortran77