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Firlakuza [10]
3 years ago
11

Trace the evaluation of the following expressions, and give their resulting values. Make sure to give a value of the appropriate

type (such as including a .0 at the end of a double or quotes around a String).
4 + 1 + 9 + "." + (-3 + 10) + 11 / 3
8 + 6 * -2 + 4 + "0" + (2 + 5)
1 + 1 + "8 - 2" + (8 - 2) + 1 + 1
5 + 2 + "(1 + 1)" + 4 + 2 * 3
"1" + 2 + 3 + "4" + 5 * 6 + "7" + (8 + 9)
Computers and Technology
1 answer:
Masja [62]3 years ago
7 0

Answer:

(a)\ 14"."43

(b)\ 0"0"7

(c)\ 2"8 - 2"611

(d)\ 7"(1 + 1)"46

(e)\ "1"23"4"30"7"17

Explanation:

Required

Evaluate each expression

The simple rules to follow are:

(1) All expressions in bracket will be evaluated based on its data type

(2) Divisions will return only integer values

(3) Integers immediately after string values will be concatenated (not added)

So, the results are as follows:

(a)\ 4 + 1 + 9 + "." + (-3 + 10) + 11 / 3

Evaluate till a string is encountered

4 + 1 + 9 = 14

Followed by "."

Then:

(-3 + 10)   = 7

11/ 3 = 3

"." + (-3 + 10) + 11/3 = "."73 -- because expressions after string operations are concatenated.

So, we have:

4 + 1 + 9 + "." + (-3 + 10) + 11 / 3 = 14"."43

(b)\ 8 + 6 * -2 + 4 + "0" + (2 + 5)

Evaluate till a string is encountered

8 + 6 * -2 + 4 =0

"0" + (2 + 5) ="0"7

So, we have:

8 + 6 * -2 + 4 + "0" + (2 + 5) = 0"0"7

(c)\ 1 + 1 + "8 - 2" + (8 - 2) + 1 + 1

Evaluate till a string is encountered

1 + 1 = 2

"8 - 2" + (8 - 2) + 1 + 1 = "8 - 2"611

So, we have:

1 + 1 + "8 - 2" + (8 - 2) + 1 + 1 = 2"8 - 2"611

(d)\ 5 + 2 + "(1 + 1)" + 4 + 2 * 3

Evaluate till a string is encountered

5 + 2 =7

"(1 + 1)" + 4 + 2 * 3 = "(1 + 1)"46 --- multiply 2 and 3, then concatenate

So, we have:

5 + 2 + "(1 + 1)" + 4 + 2 * 3= 7"(1 + 1)"46

(e)\ "1" + 2 + 3 + "4" + 5 * 6 + "7" + (8 + 9)

Since a string starts the expression, the whole expression will be concatenated except the multiplication and the expressions in bracket.

So, we have:

"1" + 2 + 3 + "4" + 5 * 6 + "7" + (8 + 9) = "1"23"4"30"7"17

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;one is the variable to save the number and other is the length

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%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

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read2:

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exitm:

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%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

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%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

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%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

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mov ebx, 1

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Explanation:

For an assembly code/language that has the conditions given in the question, the program that tests the macro, passing it operands of various sizes is given below;

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;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

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%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

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%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

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;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

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mov eax, 4

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mov eax, 1

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