Ner*_*ool 3 string ada case-statement
我试图在case语句中使用一个字符串,但它让我expected a discrete type. Found type Standard.String明白字符串不是离散的.我想知道是否有工作.这是我的代码:
function Is_Valid_Direction(Direction_To_Go : in String) return Integer is
Room : Integer := 0;
begin
--if (Direction_To_Go = "NORTH" or Direction_To_Go = "N") then
-- Room := Building(currentRoom).exits(NORTH);
--elsif (Direction_To_Go = "SOUTH" or Direction_To_Go = "S") then
-- Room := Building(currentRoom).exits(SOUTH);
--elsif (Direction_To_Go = "EAST" or Direction_To_Go = "E") then
-- Room := Building(currentRoom).exits(EAST);
--elsif (Direction_To_Go = "WEST" or Direction_To_Go = "W") then
-- Room := Building(currentRoom).exits(WEST);
--elsif (Direction_To_Go = "UP" or Direction_To_Go = "U") then
-- Room := Building(currentRoom).exits(UP);
--elsif (Direction_To_Go = "DOWN" or Direction_To_Go = "D") then
-- Room := Building(currentRoom).exits(DOWN);
--end if;
case Direction_To_Go is
when "NORTH" | "N" => Room := Building(currentRoom).exits(NORTH);
when "SOUTH" | "S" => Room := Building(currentRoom).exits(SOUTH);
when "EAST" | "E" => Room := Building(currentRoom).exits(EAST);
when "WEST" | "W" => Room := Building(currentRoom).exits(WEST);
when "UP" | "U" => Room := Building(currentRoom).exits(UP);
when "DOWN" | "D" => Room := Building(currentRoom).exits(DOWN);
when others => Room := 0;
end case;
return Room;
end Is_Valid_Direction;
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注释部分正在完成我想要的,但使用if语句.我只是想看一下案例陈述是否可行.
您可以将字符串映射到离散类型.最简单的是枚举类型:
procedure Light (Colour : in String) is
type Colours is (Red, Green, Blue);
begin
case Colours'Value (Colour) is -- ' <- magic ;-)
when Red =>
Switch_Red_LED;
when Green =>
Switch_Green_LED;
when Blue =>
Switch_Blue_LED;
end case;
exception
when Constraint_Error =>
raise Constraint_Error with "There is no " & Colour & " LED.";
end Light;
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