我正在执行与该站点上已经解决的标志包问题相似的任务（Package bodies and main programs. Simple assignment (Ada)]
但是这次我需要画一个有4折的书皮，如下所示：
Book with folds at the breaking points: 3, 10, 15, 17.
Height = 2, Width = 20,
Title (string(1..100),
Author (string(1..50).
<Title>
<Author>
|--|------|----|-|---|
| | | | | |
| | | | | |
|--|------|----|-|---|
我从exercise_bluebook_main.adb的第25行中发现了很多问题，但是最让我感到困扰的是
exercise_bluebook_main.adb：25：11：预期类型为“ Standard.Integer”，exercise_bluebook_main.adb：25：11：已找到定义为“ Breaking_Array”的类型在exercise_bluebook.ads:7
为什么当为定义的数组输入Im时，为什么期望标准整数？此外，我认为在主过程行15 Im的声明部分会遇到麻烦，因为：= get_line已被“标题”占用。任何人，您将如何解决？
也许代码中的错误更多，我要感谢任何和所有输入，只要它具有建设性！
我的.ads文件
with ada.text_IO; use ada.text_IO;
with ada.integer_text_IO; use ada.integer_text_IO;
package Exercise_BlueBook is
subtype Break_Points is integer range 1..20;
type Breaking_Array is array (1..4) of Break_Points;
type Book_Type is private;
procedure Get (Item: out Book_Type;
Title: in String;
Author: in String;
Width: in Integer;
Height: in Integer;
Break_Points: in Breaking_Array);
procedure Put (Item: in Book_Type);
private
type Book_Type is
record
Title : String(1..100);
Title_L : Integer;
Author : String(1..50);
Author_L : Integer;
Width : Integer;
Height : Integer;
Break_Points : Integer;
end record;
end Exercise_BlueBook;
我的包主体adb文件
with ada.text_IO; use ada.text_IO;
with ada.integer_text_IO; use ada.integer_text_IO;
with Exercise_Bluebook; use Exercise_Bluebook;
package body Exercise_BlueBook is
procedure Get (Item: out Book_Type;
Title: in String;
Author: in String;
Width: in Integer;
Height: in Integer;
Break_Points: in Breaking_Array) is
begin
Item.Title_L := Integer'Min (Item.Title'Length, Title'Length);
Item.Title (1..Item.Title_L) := Title(Title'First .. Item.Title_L);
Item.Author_L := Integer'Min (Item.Author'Length, Author'Length);
Item.Author (1..Item.Author_L) := Author (Author'First .. Item.Author_L);
Item.Width := Width;
Item.Height := Height;
Item.Break_Points := Break_Points;
end Get;
procedure Put (Item: in Book_Type) is
begin
Put_Line(Item.Title(1..Item.Title_L));
Put_Line(Item.Author(1..Item.Author_L));
for H in 1..Item.Height loop
Put("!");
for I in 1..Item.Width loop
Put("-");
if I = Breaking_Array(1) then
Put("!");
elsif I = Breaking_Array(2) then
Put("!");
elsif I = Breaking_Array(3) then
Put("!");
elsif I = Breaking_Array(4) then
Put("!");
end if;
end loop;
Put_Line("!");
end loop;
end Put;
end Exercise_BlueBook;
我的主要步骤
with ada.text_IO; use ada.text_IO;
with ada.integer_text_IO; use ada.integer_text_IO;
with Exercise_Bluebook; use Exercise_Bluebook;
procedure Exercise_BlueBook_Main is
B : Book_Type;
begin
Put("Enter the name of the book: ");
declare
Title : constant String := Get_Line; -- First Get_Line already taken
Author: constant String := Get_Line; -- How do I solve this for "Author"?
Width: Integer;
Height: Integer;
Break_Points : Exercise_BlueBook.Breaking_Array;
begin
Put("Enter the book's width: ");
Get(Width);
Put("Enter the book's height: ");
Get(Height);
Put("Enter the breaking points for the fold: ");
Get(Break_Points); -- Won't read this as an array?
Get(B,
Title => Title,
Author => Author,
Width => Width,
Height => Height,
Break_Points => Break_Points);
end;
New_Line;
Put(B);
end Exercise_BlueBook_Main;
Get程序包中的Ada.Integer_Text_IO子程序将只读取一个整数。您可以通过添加循环来扩展它：main.adb
with Ada.Text_IO;
with Ada.Integer_Text_IO;
procedure Main is
package TIO renames Ada.Text_IO;
package IIO renames Ada.Integer_Text_IO;
type Int_Array is array (Natural range <>) of Integer;
---------
-- Get --
---------
function Get (From : String; Max_Elems : Natural) return Int_Array is
Result : Int_Array (1 .. Max_Elems);
Elems : Natural := 0;
Last : Natural := From'First - 1;
begin
while Last + 1 <= From'Last and Elems + 1 <= Max_Elems loop
declare
Value : Integer;
begin
IIO.Get (From (Last + 1 .. From'Last), Value, Last);
Elems := Elems + 1;
Result (Elems) := Value;
end;
end loop;
return Result (1 .. Elems);
end Get;
begin
declare
Arr : Int_Array := Get (TIO.Get_Line, 10);
begin
TIO.New_Line;
TIO.Put_Line("Result:");
for Val of Arr loop
IIO.Put (Val, Width => 3);
TIO.New_Line;
end loop;
end;
exception
when TIO.Data_Error =>
TIO.Put_Line ("Invalid input.");
end Main;
输出
$ ./main
1 2 3
Result:
1
2
3
（无效输入）输出
$ ./main
1 2 x 4 5
Invalid input.