下面在Clojure中使用defprotocol和defrecord的休息参数有什么问题?
(defprotocol prot
(f [this] [this & rest]))
(defrecord rec []
prot
(f [this] "one arg")
(f [this & rest] "more than one arg"))
(prn (f (rec.)))
; (prn (f (rec.) 5))
(prn (f (rec.) 5 6))
; (prn (f (rec.) 5 6 7))
上面的代码打印出我期望的输出:
"one arg"
"more than one arg"
但如果我取消注释任何一个注释行,我会得到以下异常:
Exception in thread "main" java.lang.IllegalArgumentException: No single method: f of interface: user.prot found for function: f of protocol: prot (bug.clj:10)
at clojure.lang.Compiler.analyzeSeq(Compiler.java:5376)
at clojure.lang.Compiler.analyze(Compiler.java:5190)
at clojure.lang.Compiler.analyze(Compiler.java:5151)
at clojure.lang.Compiler$InvokeExpr.parse(Compiler.java:3057)
at clojure.lang.Compiler.analyzeSeq(Compiler.java:5371)
at clojure.lang.Compiler.analyze(Compiler.java:5190)
at clojure.lang.Compiler.analyze(Compiler.java:5151)
at clojure.lang.Compiler$BodyExpr$Parser.parse(Compiler.java:4670)
at clojure.lang.Compiler$FnMethod.parse(Compiler.java:4328)
at clojure.lang.Compiler$FnExpr.parse(Compiler.java:3173)
at clojure.lang.Compiler.analyzeSeq(Compiler.java:5367)
at clojure.lang.Compiler.analyze(Compiler.java:5190)
at clojure.lang.Compiler.eval(Compiler.java:5421)
at clojure.lang.Compiler.load(Compiler.java:5857)
at clojure.lang.RT.loadResourceScript(RT.java:340)
at clojure.lang.RT.loadResourceScript(RT.java:327)
at clojure.lang.RT.loadResourceScript(RT.java:319)
at clojure.main$load_script.invoke(main.clj:220)
at clojure.main$script_opt.invoke(main.clj:273)
at clojure.main$main.doInvoke(main.clj:354)
at clojure.lang.RestFn.invoke(RestFn.java:409)
at clojure.lang.Var.invoke(Var.java:365)
at clojure.lang.AFn.applyToHelper(AFn.java:163)
at clojure.lang.Var.applyTo(Var.java:482)
at clojure.main.main(main.java:37)
Caused by: java.lang.IllegalArgumentException: No single method: f of interface: user.prot found for function: f of protocol: prot
at clojure.lang.Compiler$InvokeExpr.<init>(Compiler.java:2880)
at clojure.lang.Compiler$InvokeExpr.parse(Compiler.java:3063)
at clojure.lang.Compiler.analyzeSeq(Compiler.java:5371)
... 24 more
为什么会这样?
答案 0 :(得分:6)
据我所知,协议不支持可变参数。可能发生的是“&amp;”被视为参数符号而不是可变指标/列表理解。