(13 Points) One way to calculate the harmonic mean of two numbers x and y is usi

Question

(13 Points) One way to calculate the harmonic mean of two numbers x and y is using the expression 3. 2xy Complete the following windows32 program to input two numbers and calculate and display their harmonic mean in dddd.dd format. The boxed areas indicate where code is missing. ; student' s name here ; program to input two numbers, i then calculate and display their harmonic mearn ; current date here 586 . MODEL FLAT INCLUDE ?o.h : header file for input/output STACK 4096 DATA DWORD 2 DWORD 2 prompt1 BYTE "First number", o prompt2 BYTE "second number", o inArea BYTE 20 DUP (2) mean BYTE11 DUP (?), 0 meanLbl BYTE "Harmonic mean" meanout BYTE 4 DUP (?),,2 DUP (2),

Explanation / Answer

as per requirement blank spaces contain required code (in bold)

and comments in program specify explanation for used instructions.

Code here
;-----------------------------------------------------------------------

.586

.MODEL FLAT

INCLUDE io.h

.STACK 4096

.DATA

x DWORD ?
y DWORD ?

prompt1 BYTE "First number", 0
prompt2 BYTE "Second number", 0
inArea BYTE 20 DUP (?)
mean BYTE 11 DUP (?), 0
meanLbl BYTE "Harmonic Mean", 0
meanOut BYTE 4 DUP(?), '.', 2 DUP (?), 0

.CODE

_MainProc Proc

    input prompt1, inArea, 20
    atod inArea
    mov x,eax

;repeat for second number

    input prompt2, inArea, 20
    atod inArea
    mov y,eax

;calculate 100*(harmonic mean) in EAX

mov eax,x   ;here mult(x*y) result in eax
mov edx, y
mul edx

mov ebx,2    ;here mult(2*(above result)) result in eax
mul eax

mov x,eb      ;here addition(x+y) result in ebx destination register
add ebx,y

fdiv ebx       ; here divison where divisor in ebx and divident in eax result in eax

mov eax,100   ; here 100*(harmonic mean) and result in eax
fmul eax

mov mean,eax

dtoa mean, eax ; convert to ASCII char

;copy digits, one at a time to meanOut

fld mean    ;load floating-point value
mov cx, 2
mov si,6

fstp ax        ;store floating-point value and pop

loop1:             
mov meanOut[si], ax
dec cx
jnz loop1

mov si,4
mov cx,4

loop2:
mov meanOut[si],ax
dec cx
jnz loop2

;output label and mean
mov dl, meanLbl
mov    ah, 02
int    21h

mov cx, 7
    l1:
    lea si, meanOut
    mov bl, meanOut[si]
    mov dl, bl

    mov ah, 2
    int 21h
    inc si
    loop l1
         

mov eax,0
ret

_MainProc ENDP

END
  

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