列表操作的数据类型
datatype for list operation
我有这个问题要解决
我已经设法解决了它，但没有那个功能 execute_ops。这是我的解决方案：
datatype list_op = insert of int*int | delete of int | setsize of int;
exception wrong_case;
fun insert(x,pos,nil) = [x]
| insert(x,pos,h::t) = if (pos=h) then x::h::t else h::insert(x,pos,t);
fun setsize(0) = nil
| setsize(1) = [0]
| setsize(n) = 0 :: setsize(n-1);
fun delete (l,0) = l
| delete (nil,_) = raise wrong_case
| delete (h::t,n) = if n = h then h::t else delete(t,n-1);
insert(3, 4, delete(setsize(5), 3));
首先我不确定我的解决方案是否正确，因为我不明白setsize是什么。
如何正确编写代码？
谢谢
因此，对于这三个操作中的每一个，您似乎都定义了 helper 函数来执行操作背后的逻辑。除了修复此逻辑使其工作之外，剩下的就是将这些函数拼凑在一起以形成函数 execute_ops.
如果这些值同时是值构造函数[=58=，则不得将函数命名为insert、delete和setsize ] 你的数据类型。简而言之，一旦您定义了数据类型，然后在同一范围内定义了函数，这些函数将遮盖值构造函数并使您无法表达类型 list_op.
重要的是，我怀疑会造成混淆，一点是 insert、delete 和 setsize 不是对列表执行操作的函数。它们是值构造函数，很像 leaf 和 node 用于制作二叉树：
datatype tree = leaf | node of tree*int*tree
在insert中比较pos=h没有意义。列表中的元素与其位置不同。您希望在 位置 pos 处插入 x，而不是在 等于 pos 的第一个元素处插入 x。一个值不一定等于它在列表中出现的位置。
不幸的是，delete完全没有意义；你的代码说的是“如果数字 n 等于列表中的元素，那么 return 整个列表。否则，删除元素并继续删除元素，直到出现这种情况，或者列表为空."
根据定义，在 delete 中引发异常也是不必要的。
我不确定你不理解 setsize 是什么意思。它应该是一个设置列表大小的操作。重复赋值文本，如果大小小于当前长度，删除多余的元素，否则用零扩展列表的末尾。
这是解决作业的模板：
datatype list_op = insert of int*int | delete of int | setsize of int
fun insert_h (0, elem, x::xs) = ... (* insert elem instead of x *)
| insert_h (pos, elem, []) = ... (* list was too short *)
| insert_h (pos, elem, x::xs) = ... (* not there yet *)
fun setsize_h (0, xs) = ... (* wonder what list has size 0 *)
| setsize_h (n, []) = ... (* extend list n times more *)
| setsize_h (n, x::xs) = ... (* not there yet *)
fun delete_h (0, x::xs) = ... (* delete element at this position *)
| delete_h (n, []) = ... (* list was too short *)
| delete_h (n, x::xs) = ... (* not there yet *)
fun execute_ops [] xs = ... (* no more operations to execute *)
| execute_ops (list_op::list_ops) xs =
let val new_xs = (case list_op of
insert (pos, elem) => insert_h (pos, elem, xs)
| delete pos => delete_h (pos, xs)
| setsize size => setsize_h (size, xs))
in execute_ops list_ops new_xs
end
您可能希望单独或组合使用 execute_ops:
来测试这些函数
val test_insert_h_1 = insert_h (2, 7, [1,2,3,4]) = [1,2,7,4]
val test_insert_h_2 = insert_h (9, 5, [1,2,3,4]) = [1,2,3,4,5]
val test_setsize_h_1 = setsize_h (2, [5,6,7,8]) = [5,6]
val test_setsize_h_2 = setsize_h (5, [1,2,3]) = [1,2,3,0,0]
val test_delete_h_1 = delete_h (3, [4,5,6,7,8]) = [4,5,6,8]
val test_delete_h_2 = delete_h (9, [1,2,3]) = [1,2,3]
val test_execute_ops_1 =
execute_ops [insert (0, 5), insert (1, 6), insert (2, 7)] [2,3,5] = [5,6,7]
val test_execute_ops_2 =
execute_ops [setsize 5, insert (4, 9)] [] = [0,0,0,0,9]
val test_execute_ops_3 =
execute_ops [setsize 6, insert (1, 5), delete 3] [8,8,8,8,9] = [8,5,8,9,0]
我有这个问题要解决
我已经设法解决了它，但没有那个功能 execute_ops。这是我的解决方案：
datatype list_op = insert of int*int | delete of int | setsize of int;
exception wrong_case;
fun insert(x,pos,nil) = [x]
| insert(x,pos,h::t) = if (pos=h) then x::h::t else h::insert(x,pos,t);
fun setsize(0) = nil
| setsize(1) = [0]
| setsize(n) = 0 :: setsize(n-1);
fun delete (l,0) = l
| delete (nil,_) = raise wrong_case
| delete (h::t,n) = if n = h then h::t else delete(t,n-1);
insert(3, 4, delete(setsize(5), 3));
首先我不确定我的解决方案是否正确，因为我不明白setsize是什么。
如何正确编写代码？
谢谢
因此，对于这三个操作中的每一个，您似乎都定义了 helper 函数来执行操作背后的逻辑。除了修复此逻辑使其工作之外，剩下的就是将这些函数拼凑在一起以形成函数 execute_ops.
如果这些值同时是值构造函数[=58=，则不得将函数命名为
insert、delete和setsize] 你的数据类型。简而言之，一旦您定义了数据类型，然后在同一范围内定义了函数，这些函数将遮盖值构造函数并使您无法表达类型 list_op.重要的是，我怀疑会造成混淆，一点是
insert、delete和setsize不是对列表执行操作的函数。它们是值构造函数，很像leaf和node用于制作二叉树：datatype tree = leaf | node of tree*int*tree在
insert中比较pos=h没有意义。列表中的元素与其位置不同。您希望在 位置pos处插入 x，而不是在 等于pos的第一个元素处插入 x。一个值不一定等于它在列表中出现的位置。不幸的是，
delete完全没有意义；你的代码说的是“如果数字 n 等于列表中的元素，那么 return 整个列表。否则，删除元素并继续删除元素，直到出现这种情况，或者列表为空."根据定义，在
delete中引发异常也是不必要的。我不确定你不理解 setsize 是什么意思。它应该是一个设置列表大小的操作。重复赋值文本，如果大小小于当前长度，删除多余的元素，否则用零扩展列表的末尾。
这是解决作业的模板：
datatype list_op = insert of int*int | delete of int | setsize of int
fun insert_h (0, elem, x::xs) = ... (* insert elem instead of x *)
| insert_h (pos, elem, []) = ... (* list was too short *)
| insert_h (pos, elem, x::xs) = ... (* not there yet *)
fun setsize_h (0, xs) = ... (* wonder what list has size 0 *)
| setsize_h (n, []) = ... (* extend list n times more *)
| setsize_h (n, x::xs) = ... (* not there yet *)
fun delete_h (0, x::xs) = ... (* delete element at this position *)
| delete_h (n, []) = ... (* list was too short *)
| delete_h (n, x::xs) = ... (* not there yet *)
fun execute_ops [] xs = ... (* no more operations to execute *)
| execute_ops (list_op::list_ops) xs =
let val new_xs = (case list_op of
insert (pos, elem) => insert_h (pos, elem, xs)
| delete pos => delete_h (pos, xs)
| setsize size => setsize_h (size, xs))
in execute_ops list_ops new_xs
end
您可能希望单独或组合使用 execute_ops:
val test_insert_h_1 = insert_h (2, 7, [1,2,3,4]) = [1,2,7,4]
val test_insert_h_2 = insert_h (9, 5, [1,2,3,4]) = [1,2,3,4,5]
val test_setsize_h_1 = setsize_h (2, [5,6,7,8]) = [5,6]
val test_setsize_h_2 = setsize_h (5, [1,2,3]) = [1,2,3,0,0]
val test_delete_h_1 = delete_h (3, [4,5,6,7,8]) = [4,5,6,8]
val test_delete_h_2 = delete_h (9, [1,2,3]) = [1,2,3]
val test_execute_ops_1 =
execute_ops [insert (0, 5), insert (1, 6), insert (2, 7)] [2,3,5] = [5,6,7]
val test_execute_ops_2 =
execute_ops [setsize 5, insert (4, 9)] [] = [0,0,0,0,9]
val test_execute_ops_3 =
execute_ops [setsize 6, insert (1, 5), delete 3] [8,8,8,8,9] = [8,5,8,9,0]