THE MICROARCHITECTURE LEVEL - Structured Computer Organization - page 277

If the IJVC opcode is DUP , POP ,or SWAP , the stack must be adjusted. The
DUP instruction simply replicates the top word of the stack. Since the value of this
word is already stored in TOS , the operation is as simple as incrementing SP to
point to the new location and storing TOS to that location. The POP instruction is
almost as simple, just decrementing SP to discard the top word on the stack. How-
ever, in order to maintain the top word in TOS it is now necessary to read the new
top word in from memory and write it into TOS . Finally, the SWAP instruction
involves swapping the values in two memory locations: the top two words on the
stack. This is made somewhat easier by the fact that TOS already contains one of
those values, so it need not be read from memory. This instruction will be dis-
cussed in more detail later.
The BIPUSH instruction is a little more complicated because the opcode is fol-
lowed by a single byte, as shown in Fig. 4-18. The byte is to be interpreted as a
signed integer. This byte, which has already been fetched into MBR in Main1 , must
be sign-extended to 32 bits and pushed onto the top of the stack. This sequence,
therefore, must sign-extend the byte in MBR to 32 bits, and copy it to MDR . Finally,
SP is incremented and copied to MAR , permitting the operand to be written out to
the top of stack. Along the way, this operand must also be copied to TOS . Note
that before returning to the main program, PC must be incremented and a fetch op-
eration started so that the next opcode will be available in Main1 .

BIPUSH
(0

BYTE

10)

×

Figure 4-18. The BIPUSH instruction format.

Next consider the ILOAD instruction. ILOAD also has a byte following the op-
code, as shown in Fig. 4-19(a), but this byte is an (unsigned) index to identify the
word in the local variable space that is to be pushed onto the stack. Since there is
only 1 byte, only 2 8 256 words can be distinguished, namely, the first 256 words

in the local variable space. The ILOAD instruction requires both a read (to obtain
the word) and a write (to push it onto the top of the stack). In order to determine
the address for reading, however, the offset, contained in MBR , must be added to
the contents of LV . Since both MBR and LV can be accessed only through the B bus,
first LV is copied into H (in iload1 ), then MBR is added. The result of this addition
is copied into MAR and a read initiated (in iload2 ).

=

ILOAD
(0x15)

WIDE
(0xC4)

ILOAD
(0x15)

INDEX
BYTE 1

INDEX
BYTE 2

INDEX

(b)

(a)

Figure 4-19. (a) ILOAD with a 1-byte index. (b) WIDE ILOAD with a 2-byte index.