如何将行读取为 SIMPLE-BASE-STRING？

Question

所以，下面的代码

(defun read-8bit-lines (path)
  (with-open-file (stream path :element-type 'base-char :external-format :ascii)
    (loop
      for line = (read-line stream nil)
      while line
      collect line)))

以 SIMPLE-ARRAY-CHARACTER 类型的字符串列表的形式读入文件。我想要 SIMPLE-BASE-STRING 代替。我可以用 coerce:

(defun read-8bit-lines (path)
  (with-open-file (stream path :element-type 'base-char :external-format :ascii)
    (loop
      for line = (read-line stream nil)
      while line
      collect (coerce line 'simple-base-string))))

有没有不用求助于coerce的方法？我是运行 sbcl 2.0.5.

Answer 1

便携，没有。实现可以自由 return 他们认为合理的任何类型的字符串。

我会使用一个简单的包装函数，并可能声明它内联：

(declaim (inline read-simple-line))
(defun read-simple-line (stream)
  (some-> (read-line stream nil)
          (coerce 'simple-base-string)))

（Some-> 来自 arrows）。

如何将行读取为 SIMPLE-BASE-STRING？

How to read lines as SIMPLE-BASE-STRING?

lisp

sbcl

common-lisp