Question

使用guile 1.8或guile 2，下面的代码读取过去的EOF，看似多了几行，然后停止。这在提取的较大程序中的效果似乎是破坏了先前读取的数据。我是否错误地使用了读取线或测试eof-object？

(use-modules (ice-9 rdelim))

(define f
  (lambda (p)
    (let loop ((line (read-line p)))
      (format #t "line: ~a\n" line)
      (if (not (eof-object? (peek-char p)))
      (begin
        (let ((m (string-match "^[ \t]*#" line)))
          (if m
          (begin
              (format #t "comment: ~a\n" (match:string m))
              (loop (read-line p))
            )))
        (format #t "read next line\n")
        (loop (read-line p)))))))

(define main
  (lambda ()
    (let ((h (open-input-file "test")))
      (f h))))

这是一个最小样本虚拟输入文件：

1
2
3
# comment line
4
5
1
2
3
# comment line
4
5
1
2
3
# comment line
4
5

要显示问题需要多于几行。请注意代码示例的长度，但只有在代码达到这种复杂程度时才会出现问题（尽管很小）。

Answer 1

我建议重写该过程，它似乎不是正确的读取文件并循环其行的方法。请尝试这个：

(define (f)
  (let loop ((line (read-line)))
    (if (not (eof-object? line))
        (begin
          (format #t "line: ~a\n" line)
          (let ((m (string-match "^[ \t]*#" line)))
            (if m (format #t "comment: ~a\n" line)))
          (format #t "read next line\n")
          (loop (read-line))))))

(define (main)
  (with-input-from-file "test" f))

使用您的示例输入，调用(main)在控制台上输出以下输出，这有望达到预期目的：

line: 1
read next line
line: 2
read next line
line: 3
read next line
line: # comment line
comment: # comment line
read next line
line: 4
read next line
line: 5
read next line
line: 1
read next line
line: 2
read next line
line: 3
read next line
line: # comment line
comment: # comment line
read next line
line: 4
read next line
line: 5
read next line
line: 1
read next line
line: 2
read next line
line: 3
read next line
line: # comment line
comment: # comment line
read next line
line: 4
read next line
line: 5
read next line

Guile Scheme阅读过去EOF的阅读

1 个答案: