Question

我有这段代码：

(def x 10)

(def io (java.io.StringWriter.))

(with-redefs [x 11]
  (with-redefs [io (java.io.StringWriter.)]
    (.write io "one"))
  (with-redefs [io (java.io.StringWriter.)]
    (.write io "two")
    (.toString io)))

(with-redefs [x 11]
  (with-redefs [*out* (java.io.StringWriter.)]
    (print "one"))
  (with-redefs [*out* (java.io.StringWriter.)]
    (print "two")
    (.toString *out*)))

第一个with-redefs块返回＆＃34;两个＆＃34;正如所料。第二个返回＆＃34; onetwo＆＃34;。就好像它出于某种原因重新使用同样的模拟**一样。这是预期的行为吗？

Answer 1

问题在于，您关心的*out*是线程本地的，而不是根值。尝试：

(with-redefs [x 11]
  (binding [*out* (java.io.StringWriter.)]
    (print "one"))
  (binding [*out* (java.io.StringWriter.)]
    (print "two")
  (.toString *out*)))

with-redefs宏仅改变根绑定;任何线程本地绑定都不受影响。当代码由REPL服务器执行时，它是在将*out*重新绑定到客户端特定的编写器的上下文中完成的，这样就可以捕获输出并将其发送回客户端以供显示（否则它会转到服务器的标准输出。如果从命令行而不是REPL运行，原始代码可能会起作用。

嘲笑中的棘手行为* out *

1 个答案: