68K Subroutine by trigyy

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This article will explain the concepts of subroutine and its application. The term Nested Subroutine also will be explained in this article. Lastly, what is the delay in subroutine and how to program in 68000 microprosessor

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									        Subroutine
Concepts and Applications




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Objectives
 To examine the concepts of subroutine and
 its application




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Outcome
To be able to write a delay subroutine
To be able to compute the duration of a
subroutine
To be able to tackle large programs




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Subroutine

Many programs require certain instructions to
performed several times.
The required instructions can be put into a
subroutine.
subroutine
A subroutine is a sequence of instructions that
is written just once but executed when needed
by the main program.
  Example: Time delay routine.

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Main program



                                Subroutine




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 In a 68000 system the subroutine is
 called/invoke by the BSR instruction.
 The RTS instruction is used to return to
 the main program.
   RTS is the last instruction in the subroutine.


SAQ
      Why do we need to use subroutine?



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Reasons For Using Subroutines
Advantages
 Programs are usually shorter
 Programs are more reliable
   Once a subroutine has been developed and
   tested it can be re-used without further testing
 Programs are more maintainable because
 they are shorter, simpler and more
 readable. Also it is easy to change the
 functionality of a subroutine, if required.

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Disadvantages
  The execution speed of a program is
  slower if it uses subroutines
  At machine code level, there are some
  complications introduced e.g. saving the
  environment of the calling program.
  Difficult program bugs can occur due to
  subroutine side effects e.g. stack,
  memory corruption's due to invalid
  parameter passing register saving


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Saving the CPU Environment
  In some cases, when a subroutine is
  called, it will need to use the CPU
                                                   ?
  registers.
  Before a subroutine is executed, the
  registers must be saved because their
  contents belong to the main program
  and will probably be needed when the
  subroutine returns.                              CPU registers –
    Main program saves any registers it needs      (Environment)
    onto stack and recovers them after the
    subroutine call.
    Subroutine responsible for saving registers.

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Nested Subroutine
A nested subroutine is a subroutine that
calls another subroutine.




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Calculating Time Delays
 For a CPU running at a constant clock rate:
     Cycle Time = 1 / clock frequency
 Every machine or assembly instruction takes one or
 more clock cycles to complete.
 Consider the following time delay routine:

                  move.b              #value,d0
    loop          sub.b               #1,d0
                  bne                 loop
   The delay will depend upon the variable value



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From the Table D-5, “Immediate Instruction
Execution Times” - sub.b #1,d0 has an execution
time of 8 cycles
                                     No. of read cycles
                    4(1/0)
   No. of clock                           No. of write cycles
   periods
From the Table D-9, “Conditional and Branch
Instruction Execution Times” - bne loop has an
execution time of 10 cycles (bne is one of the Bcc
group)
Each pass through the loop will take 4 + 10 = 14
cycles.

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The Flite 68K system has a 10MHz clock, giving a
time for each cycle of 0.0000001s or 0.1µs Recall
that delay loop takes 14 cycles for each pass. This
represents a time delay of 14 x 0.1 µs = 1.4 µs.
Now, if the variable value is FFh (255 decimal).
The delay is:
 255 x 1.4 µS = 0.000357 S = 357 µS (0.357 ms)
Now, if the variable value is FFFFh (65535
decimal). The delay is:
   65535 x 1.4 µS = 0.091749 S = 91.749 ms



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Worked Example:
Write a program fragment which gives a delay of
115 mS.

Solution:
    115 mS
           = 82142.8 = 82143
    1.4 µS

  So 82143 is the value to be loaded into the D0
  register.
    The source code is as follows;


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       move.b                #82142,d0
loop   sub.b                 #1,d0
       bne                   loop




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Compute the total execution time for the
following 68000 assembler code fragment


             Move.b         #0A,D1
     Here    Add.w          (A1)+,D2
             Subq.b          #1,D1
             BNE             Here




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           Instructions                    Clock cycles
                                        Overhead    Loop
        Move.b    #10,D1                    8
Here    Add.w     (A1)+,D2                         4
        Sub.q     #1,D1                            4
        BNE       Here                             10 (8)
                                       Loop taken   4+4+10 = 18
                                       Loop not taken 4+4+8 = 16

       4 + 9(18) + 16 = 226 clock cycles
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   Subroutines
 TACKLING A LARGE
PROGRAMMING TASK



                      Antonakos, pp.166 - 172
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Break the program down into smaller
modules or tasks.
 Each task can be thought as a subroutine.
 The subroutine can be called by the main
 program when needed.
 Each subroutine performs a specific task and
 hence each subroutine can be easily tested and
 corrected.


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Testing the Modules
 Testing of each subroutine module can be
 done separately, through a special program
 called a driver.
          driver
  The driver supplies the subroutine with sample
  input data and examines the sub-routine's output
  for correctness.
  It is up to the programmer to select the type and
  quantity of the sample data.


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Test Data

            Test Driver

             Module
             Under
              Test



                                  Results




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Subroutine to be tested
                                         Antonakos, pp.171


The following subroutine was written to solve the
quadratic equation
             y = 5x2 - 2x + 6

Quad   move.w   D0,D1   ;Save copy of input value
       muls     D0,D1   ;Compute X^2
       muls     #5,D1   ;Compute 5X^2
       muls     #2,D0   ;Compute 2X
       sub.w    D0,D1   ;Compute 5X^2 - 2X
       add.w    #6,D1   ;Compute 5X^2 - 2X + 6
       rts

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Determine what the correct results
should be. For example, consider the
following input and output pairs

      X – input                      y – output
           0                              6
           1                              9
          10                            486
         100                          49806

                      See pdf file for software solution

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•The software driver will send each x-input value to the
procedure one at a time and check for a match with the
expected y-output value each time.

•If all four tests pass, the driver assumes the new
routine is acceptable. If anyone test fails, an error
message is output.


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