In Ada, how do I get the current time in seconds as a value I can print out and perform operations with? - time

Question

I am doing a project for an operating systems class. I need to write a program that prints out the current time every ten seconds but also accounts for the delay of the overhead so that it does not drift when it has been running for a long time. I need it to be up to at least 1 decimal place as well.
I am stuck on step 1 as I can't figure out how to get the current time in seconds as a value. I have searched but could only find out how to get the current time in the HH:MM:SS format.
Thanks

Answer 1

Here's what I came up with:
writing_test.ads
package Writing_Test is
protected Writer is
entry write( Text : String; New_Line : Boolean:= True );
end Writer;
task Timer is
entry Start;
entry Pause;
entry Stop;
end Timer;
private
Timer_Frequency : constant Duration:= 10.0;
end Writing_Test;
writing_test.adb
with
Ada.Calendar,
Ada.Text_IO;
package body Writing_Test is
protected body Writer is
entry write( Text : String; New_Line : Boolean:= True ) when True is
begin
Ada.Text_IO.Put( Text );
if New_Line then
Ada.Text_IO.New_Line;
end if;
end;
end Writer;
task body Timer is
Active,
Stop_Task : Boolean:= False;
Next_Time : Ada.Calendar.Time;
use type Ada.Calendar.Time;
begin
MAIN:
loop
if not Active then
select
accept Start do
Active:= True;
Next_Time:= Ada.Calendar.Clock + Timer_Frequency;
end Start;
or
terminate;
end select;
else
select
accept Pause do
Active:= False;
end Pause;
or
accept Stop do
Stop_Task:= True;
end Stop;
or
delay until Next_Time;
end select;
exit MAIN when Stop_Task;
if Active then
declare
Use Ada.Calendar;
Now : Time renames Clock;
Str : String renames
Day_Duration'Image( Ada.Calendar.Seconds(Now) );
--' Formatter-correction trick
begin
Writer.write(Text => Str);
Next_Time:= Next_Time + Timer_Frequency;
end;
end if;
end if;
end loop MAIN;
end Timer;
end Writing_Test;

Answer 2

Here is a simple program that does the "print every 10s" part of the job. You could easily use this code in a separate package.
with Ada.Text_IO;
with Ada.Calendar;
procedure Periodic_Printer is
task type My_Printer_Task is
end My_Printer_Task;
task body My_Printer_Task is
use Ada.Calendar; -- for the "-" and "+" operations on Time
Start_Time : Ada.Calendar.Time;
Next_Cycle : Ada.Calendar.Time;
Period : constant Duration := 10.0;
begin
Start_Time := Ada.Calendar.Clock;
Next_Cycle := Start_Time;
loop
Ada.Text_IO.Put_Line(Duration'Image(Ada.Calendar.Clock - Start_Time)); --'
-- You could use Next_Cycle instead of Ada.Calendar.Clock - Start_Time
-- so the printing does not depend of the time needed to do the elapsed
-- time calculation
delay 3.0; -- like a long operation, takes time......
-- This pattern assumes the each cycle last less than Period
-- If you cannot ensure that, you should consider improving
-- the pattern or reduce the computation load of each cycle
Next_Cycle := Next_Cycle + Period;
delay until Next_Cycle;
end loop;
end My_Printer_Task;
Printer : My_Printer_Task;
begin
delay 90.0; -- You can do your 'real work' here.
-- Unclean way to terminate a task, you should consider improve it for a
-- real world scenario
abort Printer;
Ada.Text_IO.Put_Line("End of program");
end Periodic_Printer;