CS 2734 Computer Organization II

• Informal quiz 3 - a two-stage pipelined machine cycle.

• Review of the calling frame mechanism.

• Automatic arrays are laid out on the stack in the same way as simple variables, but upside down. Let's translate the following:

         int myfun2(void)
         {
           int i;
           int a[4];
           for (i = 0; i < 4; i++) 
               a[i] = i + 3;
           return a[3];
         }
  

  We'll place i at %fp - 4 and a[0] through a[3] in locations %fp - 20 through %fp - 8 respectively.

            .global myfun2
     myfun2:
            save %sp, -88, %sp
  	  st   %g0, [%fp - 4]    ! Set i to 0 
            sub  %fp, 20, %l0      ! Put address of a[0] in %l0 
     top_loop:
            ld   [%fp - 4], %l1    !
            subcc  %l1, 4, %g0     ! if (i - 4) >= 0, goto done
            bge   done_loop
            nop
            add   %l1, 3, %l2
            st    %l2, [%l0]       ! a[i] = i + 3
            add   %l0, 4, %l0      ! update pointer to a
            add   %l1, 1, %l1      ! i++ 
            st    %l1, [%fp - 4]
            ba    top_loop
            nop
      done_loop:
            ld   [%fp - 8], %i0    ! return a[3] 
            ret
            restore
  

  The C compiler translated this as myfun2.s. If you're curious you can compare this to the result of compiler optimization by cc myfun2.c -O -S.

• Practice with array addressing.