Ada初学者堆栈程序

问题描述 投票:2回答:2

[基本上,我有2个文件(.adb和.ads)。我是Ada的新手,还介绍了如何编译2个文件。该程序是一个基本的堆栈实现。编译.adb文件时出现此编译错误。

$ gcc -c test_adt_stack.adb
abstract_char_stack.ads:22:01: end of file expected, file can have only one compilation unit

我拥有的2个文件是:abstract_char_stack.ads

-----------------------------------------------------------
package Abstract_Char_Stack is
  type Stack_Type is private;
  procedure Push(Stack : in out Stack_Type;
                 Item  : in Character);
  procedure Pop (Stack : in out Stack_Type;
                 Char  : out Character);
private
  type Space_Type is array(1..8) of Character;
  type Stack_Type is record
    Space : Space_Type;
    Index : Natural := 0;
  end record;
end Abstract_Char_Stack;
-----------------------------------------------------------
package body Abstract_Char_Stack is
----------------------------------------------
  procedure Push(Stack : in out Stack_Type;
                  Item : in Character) is
  begin
    Stack.Index := Stack.Index + 1;
    Stack.Space(Stack.Index) := Item;
  end Push;
--------------------------------------------
  procedure Pop (Stack : in out Stack_Type;
                 Char  : out Character) is
  begin
    Char := Stack.Space(Stack.Index);
    Stack.Index := Stack.Index - 1;
  end Pop;
--------------------------------------------
end Abstract_Char_Stack;

另一个是test_adt_stack.adb

-----------------------------------------------------------
with Ada.Text_IO; use Ada.Text_IO;
with Abstract_Char_Stack; use Abstract_Char_Stack;
procedure Test_ADT_Stack is
  S1 : Stack_Type;
  S2 : Stack_Type;
  Ch : Character;
begin
  Push(S1,'H'); Push(S1,'E');  
  Push(S1,'L'); Push(S1,'L');
  Push(S1,'O');                          -- S1 holds O,L,L,E,H

  for I in 1..5 loop
    Pop(S1, Ch);  
    Put(Ch);                             -- displays OLLEH
    Push(S2,Ch); 
  end loop;                              -- S2 holds H,E,L,L,O

  New_Line;
  Put_Line("Order is reversed");

  for I in 1..5 loop
    Pop(S2, Ch);
    Put(Ch);                             -- displays HELLO
  end loop;

end Test_ADT_Stack;
-----------------------------------------------------------

我在做什么错?我只想让它编译并显示它应该执行的操作。这是研究程序分配的一种方法。但是我无法使其编译或不知道我是否做对了。

stack ada abstract-data-type gnat
2个回答
8
投票

问题在于,GNAT [和FSF GNAT是GCC使用的IIRC,不允许在单个文件中使用多个编译单元。 (这是由于他们如何管理库,但是对于初学者而言,这可能太详细了。)

解决方案,每个都需要自己的文件:

  • Abstract_Char_Stack规范(abstract_char_stack.ads
  • Abstract_Char_Stack主体(abstract_char_stack.adb
  • Test_ADT_Stack [过程]正文(test_adt_stack.adb

0
投票
why this bubble sort is not working?


with Ada.Text_Io;
use Ada.Text_Io;
generic
 type Integer is private;
 type Integer_array is Array (Natural range <>) of Integer;
procedure Bubble_Sort (A : in out Integer_Array);

procedure Bubble_Sort (A : in out Integer_Array) is
    Temp : Integer;
begin
    for I in reverse A'Range loop
       for J in A'First .. I loop
          if A(I) < A(J) then
             Temp := A(J);
             A(J) := A(I);
             A(I) := Temp;
          end if;
       end loop;
    end loop;
end Bubble_Sort;
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