ASSEMBLY LANGUAGE PROGRAMMING - Structured Computer Organization - page 730

AH and the value 2 being loaded into AL . Then line 5 adds AL to AH , making AH
equal to 3. On line 6, the contents of the variable times (10) are copied into CX .
On line 7, the address of the variable muldat , which is 2 because it is at the second
byte of the DATA segment, is loaded into BX . This is the instant in time at which
the dump of Fig. C-13(b) was made. Note that AH is 3, AL is 2, and AX is 770,

which is to be expected, as 3 × 256 + 2 = 770.

start: ! 3

CS: 00 DS=SS=ES:002
AH:03 AL:02 AX: 770
BH:00 BL:02 BX: 2
CH:00 CL:0a CX: 10
DH:00 DL:00 DX: 0
SP: 7fe0 SF O DS Z C
BP: 0000 CC - > p - -
S I : 0000 IP:0009:PC
DI : 0000 start + 4

CS: 00 DS=SS=ES:002
AH:38 AL:80 AX: 14464
BH:00 BL:02 BX: 2
CH:00 CL:04 CX: 4
DH:00 DL:01 DX: 1
SP: 7fe0 SF O DS ZC
BP: 0000 CC v > p - c
S I : 0000 IP:0011:PC
DI : 0000 start + 7

MOV AX,258 ! 4

ADDB AH,AL ! 5

MOV CX,(times) ! 6

MOV BX,muldat ! 7

MOV AX,(BX) ! 8

llp: MUL 2(BX) ! 9

LOOP llp ! 10

.SECT .DATA ! 11

times: .WORD 10 ! 12

muldat :.WORD 625, 2 ! 13

(a) (b) (c)

Figure C-13. (a) Part of a program. (b) The tracer register window after line 7
has been executed. (c) The registers.

The next instruction (line 8) copies the contents of muldat into AX . Thus, after
the return key is hit, AX will be 625.
We are now ready to enter a loop that multiplies the contents of AX by the word
addressed by 2BX (i.e., muldat + 2 ), which has the value 2. The implied destina-
tion of the MUL instruction is the DX : AX long register combination. In the first
iteration of the loop, the result fits in one word, so AX contains the result (1250),
and DX remains 0. The contents of all the registers after 7 multiplications are

shown in Fig. C-13.
Since AX started at 625, the result after those seven multiplications by 2 is
80,000. This result does not fit in AX , but the product is held in the 32-bit register
formed by the concatenation of DX : AX ,so DX is 1 and AX is 14,464. Numerically,
this value is 1

65,536 + 14,464, which is, indeed, 80,000. Note that CX is 4 here,
because the LOOP instruction decrements it every iteration. Because it started at
10, after seven executions of the MUL instruction (but only six iterations of the
LOOP instruction) we have CX set to 4.
In the next multiplication, trouble crops up. Multiplication involves AX but not
DX ,sothe MUL multiples AX (14464) by 2 to get 28,928. This results in AX being
set to 28,928 and DX being set to 0, which is numerically incorrect.

×

C.8.3 Call Command and Pointer Registers

The next example, vecprod.s is a small program that computes the inner prod-
uct of two vectors, vec1 and vec2 . It is listed in Fig. C-14.