Recursion 在clojure中递归无限,不返回值
我对clojure不熟悉。在这里您可以看到函数fp_fun。我正在使用这个函数,它将变为无穷大。我认为它没有返回值。请告诉我答案。提前谢谢Recursion 在clojure中递归无限,不返回值,recursion,clojure,clojurescript,maze,Recursion,Clojure,Clojurescript,Maze,我对clojure不熟悉。在这里您可以看到函数fp_fun。我正在使用这个函数,它将变为无穷大。我认为它没有返回值。请告诉我答案。提前谢谢 (defn fp\u fun ([x y行列向量] (如果(或(=y列)) 0) (如果(=向量[xy]“@”) 1) (如果(=vec[xy]“#”) 0) (def x1(+x1)) (def y2(-y 1)) (如果(((fp_fun x y2行列向量))1) 1) (如果(((fp_fun x1 y行列向量))1) 1) 1) (println“!
(defn fp\u fun
([x y行列向量]
(如果(或(=x行)(>=y列))
0)
(如果(=向量[xy]“@”)
1)
(如果(=vec[xy]“#”)
0)
(def x1(+x1))
(def y2(-y 1))
(如果(((fp_fun x y2行列向量))1)
1)
(如果(((fp_fun x1 y行列向量))1)
1)
1)
(println“!”x y)
0
) )
首先,我将应用emacs自动格式化程序,然后让我们研究一下这里每个表达式的含义
(defn fp_fun
([x y row col vec]
(if (or (< x 0)
(< y 0)
(>= x row)
(>= y col))
0)
(if (= vec[x y] "@")
1)
(if (= vec[x y] "#")
0)
(def x1 (+ x 1))
(def y2 (- y 1))
(if (= ( (fp_fun x y2 row col vec)) 1 )
1)
(if (= ( (fp_fun x1 y row col vec)) 1 )
1)
1)
(println "!" x y)
0) )
第二种形式是if,当有多种方法使用不同数量的参数调用此函数时
(defn
([one-arg]
result-expression-here)
([fist-arg second-arg]
other-result-expression-here))
您所拥有的是两者的一部分,因此让我们删除最后两个表达式:
(println "!" x y)
及
这就给我们留下了
(defn fp_fun
;; this function always takes 5 args, so prefer the basic form
([x y row col vec]
;; this first expression does absolutly nothing and it totally ignored
(if (or (< x 0)
(< y 0)
(>= x row)
(>= y col))
0)
;; this next expression is also ignored, only the last value in a function determines it's result
(if (= vec[x y] "@")
1)
;; so this one does nothing as well
(if (= vec[x y] "#")
0)
;; these two define some global variables
;; using def inside a function is almost always better done with a let expression
(def x1 (+ x 1))
(def y2 (- y 1))
;; this will always run because none of the above statements cause the function to
;; stop and return it's value early.
(if (= ( (fp_fun x y2 row col vec)) 1 )
1)
;; and then the stack will overflow before we get here
(if (= ( (fp_fun x1 y row col vec)) 1 )
1)
1))
您好,谢谢您的回答,但如果我添加4条if语句,它再次进入for循环会怎么样。你能告诉我吗<代码>(defn find_path[xy row col board](if(或(
0
(defn fp_fun
;; this function always takes 5 args, so prefer the basic form
([x y row col vec]
;; this first expression does absolutly nothing and it totally ignored
(if (or (< x 0)
(< y 0)
(>= x row)
(>= y col))
0)
;; this next expression is also ignored, only the last value in a function determines it's result
(if (= vec[x y] "@")
1)
;; so this one does nothing as well
(if (= vec[x y] "#")
0)
;; these two define some global variables
;; using def inside a function is almost always better done with a let expression
(def x1 (+ x 1))
(def y2 (- y 1))
;; this will always run because none of the above statements cause the function to
;; stop and return it's value early.
(if (= ( (fp_fun x y2 row col vec)) 1 )
1)
;; and then the stack will overflow before we get here
(if (= ( (fp_fun x1 y row col vec)) 1 )
1)
1))
(defn fp_fun [x y row col vec]
(if (or (< x 0)
(< y 0)
(>= x row)
(>= y col))
0
(if (= (vec x y) "@")
1
(if (= (vec x y) "#")
0
(let [x1 (+ x 1)
y2 (- y 1)]
(if (= (fp_fun x y2 row col vec) 1)
1
(if (= (fp_fun x1 y row col vec) 1)
1)))))))
(cond
(or (< x 0) (< y 0) (>= x row) (>= y col)) 0
(= (str(get-in board [x y])) "@") 1
(= (str (get-in board [x y])) "#") 0
(= (find_path x (- y 1) row col board) 1) 1
(= (find_path (+ x 1) y row col board) 1) 1
(and (>= x 0) (< y col)) (if (= (find_path x (+ y 1) row col board) 1)
1
0)
:defautl nil) ;; i'm not sure you intended to default to nil