Write code using the Microsoft Assembly Language MASM. Your program must be prop

Question

Write code using the Microsoft Assembly Language MASM. Your program must be properly Include comment lines at the end of each line. A. Integer Expression Calculation Write a program that calculate the following expression, using registers: A (A +B) - (C - D) Assign integer values to the EAX, EBX, ECX, EDX registers. Provide a screenshot or copy the values of the results and include with your program. B. Machine Language Listing Generate a listing of the above program and write a description of the machine code bytes generated for each instruction. You may have to guess at some meanings of the byte values.

Explanation / Answer

1. Program to calculate the following Exapression

a=(a+b)-(c-d)

.data
a: .word 0x42
b: .word 0x43
c: .word 0x44
d: .word 0x45

.text
start: set a, %r1
ld [%r1], %r2 ! $a$ --> %r2
set b, %r1
ld [%r1], %r3 ! $b$ --> %r3
set c, %r1
ld [%r1], %r4 ! $c$ --> %r4
set d, %r1
ld [%r1], %r5 ! $d$ --> %r5

add %r2, %r3, %r2 ! $a+b$ --> %r2
sub %r4, %r5, %r3 ! $c-d$ --> %r3
sub %r2, %r3, %r2 ! $(a+b)-(c-d)$ --> %r2
set a, %r1
st %r2, [%r1] ! $(a+b)-(c-d)$ --> a

end: ta 0

2.description of the machine code bytes generated for each instructions.

Example]::
================
The MIPS architecture provides a specific example for a machine code whose instructions are always 32 bits long.
The general type of instruction is given by the op (operation) field, the highest 6 bits.
J-type (jump) and I-type (immediate) instructions are fully specified by op. R-type (register) instructions include an
additional field funct to determine the exact operation. The fields used in these types are:

6 5 5 5 5 6 bits
[ op | EAX | EBX | ECX |EDX| funct] R-type
[ op | EAX | EBX | address/immediate] I-type
[ op | target address ] J-type

EAX, EBX, ECX,and EDX indicate register operands; shamt gives a shift amount; and the address or immediate fields contain an operand
directly.

For example, adding the registers 1 and 2 and placing the result in register 6 is encoded:

[ op | EAX | EBX | ECX |EDX| funct]
0 1 2 6 0 32 decimal
000000 00001 00010 00110 00000 100000 binary
Load a value into register 8, taken from the memory cell 68 cells after the location listed in register 3:

[ op | EAX | EBX | address/immediate]
35 3 8 68 decimal
100011 00011 01000 00000 00001 000100 binary
Jumping to the address 1024:

[ op | target address ]
2 1024 decimal
000010 00000 00000 00000 10000 000000 binary

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