有没有办法专门从数组 Java 中挑选出元素到 Ada

Question

所以我正在尝试转换这个 java 代码：

 static int evenPosition(int [] num){
        int sum = 0; 
        
        for(int i = 0; i < num.length; i++){
            sum = (num[0] + num[2] + num[4] + num[6] + num[8] + num[10])*3;
            //System.out.println(sum[i]);
        }
       
        return sum;
    }

到 ada 代码：

   type numSize is range 0 .. 9999999999;
   type SIZE is range 1 .. 12;
   type barcode is array(SIZE) of numSize;
   type odd is range 1..3;
   N1 : barcode := (0, 7, 9, 4, 0, 0, 8, 0, 4, 5, 0, 1);
   N2 : barcode := (0, 2, 4, 0, 0, 0, 1, 6, 2, 8, 6, 0);
   N3 : barcode := (0, 1, 1, 1, 1, 0, 8, 5, 6, 8, 0, 7);
   N4 : barcode := (0, 5, 1, 0, 0, 0, 1, 3, 8, 1, 0, 1);

   procedure evenPosition(num : in numSize) is
      sum : numSize := 0;
   begin
      
      sum := num;
      
        for i in odd loop
            sum := sum + sum;
            Put_Line("This is the sum");
            Put_Line(sum);
        end loop;


   end;

我对 ADA 很陌生，我的问题是有一种方法可以选择特定元素，例如 java 有什么

sum = (num[0] + num[2] + num[4]...)

如果我的措辞不正确，我很抱歉，我试图在网上找到不同的方法来做到这一点，但似乎无法找到我正在寻找的答案。请善待;-;

编辑

我很抱歉造成混乱，但我设法解决了我想做的事情。

procedure Hw2Bnguyen is
   type numSize is range 0 .. 9999999999;
   type SIZE is array(1 .. 12) of Integer;
   --type barcode is array(SIZE) of numSize;
  -- type odd is range 1..3;
  -- N1 : barcode := (0, 7, 9, 4, 0, 0, 8, 0, 4, 5, 0, 1); valid
   --N2 : barcode := (0, 2, 4, 0, 0, 0, 1, 6, 2, 8, 6, 0); not
--   N3 : barcode := (0, 1, 1, 1, 1, 0, 8, 5, 6, 8, 0, 7); valid
  -- N4 : barcode := (0, 5, 1, 0, 0, 0, 1, 3, 8, 1, 0, 1); not
  barcode : SIZE;

   num : Integer := 0;
   num2 : Integer := 0;
   num3 : Integer := 0;
   num4 : Integer := 0;
begin
   begin
     Put("Enter message: (press enter each time)");
      for i in 1..12 loop 
         barcode(i) := Integer'Value(Get_Line);
      end loop;
      
      --for k in 1..SIZE'Length loop
      --   Put(barcode(k));
      --end loop;
      
      for k in 1 ..1 loop
         num := barcode(k);
      end loop;
            for k in 3 ..3 loop
         num := num + barcode(k);
      end loop;
            for k in 5 ..5 loop
         num := num + barcode(k);
      end loop;
            for k in 7 ..7 loop
         num := num + barcode(k);
      end loop;
            for k in 9 ..9 loop
         num := num + barcode(k);
      end loop;
     
      Put("This is an odd sum: ");Put(num*3);
      New_Line;

      
      for k in 2 ..2 loop
         num2 := barcode(k);
      end loop;
      for k in 4 ..4 loop
         num2 := num2 + barcode(k);
      end loop;
      for k in 6 ..6 loop
         num2 := num2 + barcode(k);
      end loop;
      for k in 8 ..8 loop
         num2 := num2 + barcode(k);
      end loop;
      for k in 10 ..10 loop
         num2 := num2 + barcode(k);
      end loop;
      Put("This is an even sum: ");Put(num2);
      New_Line;
      
      num3 := (num*3) + num2;
      Put("Total from both numbers: "); Put(num3);

Answer 1

您可以使用mod（或rem，它们仅对负数不同）运算符来检查数字是奇数还是偶数：

procedure Hw2Bnguyen is
   type SIZE is array(1 .. 12) of Integer;
   Barcode : SIZE;

   Sum_Odd : Integer := 0;
   Sum_Even : Integer := 0;
   Sum_All : Integer := 0;
begin
   Put("Enter message: (press enter each time)");
   for Number of Barcode loop 
      Number := Integer'Value(Get_Line);
   end loop;
      
   for k in Barcode'Range loop
      if k mod 2 = 0 then
         Sum_Even := Sum_Even + Barcode(k);
      else
         Sum_Odd := Sum_Odd + Barcode(k);
      end if;
   end loop;
     
   Put("This is an odd sum: ");Put(Sum_Odd*3);
   New_Line;

   Put("This is an even sum: ");Put(Sum_Even);
   New_Line;
      
   Sum_All := (Sum_Odd*3) + Sum_Even;
   Put("Total from both numbers: "); Put(Sum_All);
end Hw2Bnguyen;