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Product Engineer II at Cadence Design Systems

Hello Dear Readers, Cadence Design Systems has a vacancy for a Product Engineer II role. Cadence is a pivotal leader in electronic design, building upon more than 30 years of computational software expertise. The company applies its underlying Intelligent System Design strategy to deliver software, hardware and IP that turn design concepts into reality.  Cadence customers are the world’s most innovative companies, delivering extraordinary electronic products from chips to boards to systems for the most dynamic market applications including consumer, hyperscale computing, 5G communications, automotive, aerospace industrial and health. The Cadence Advantage: The opportunity to work on cutting-edge technology in an environment that encourages you to be creative, innovative, and to make an impact. Cadence’s employee-friendly policies focus on the physical and mental well-being of employees, career development, providing opportunities for learning, and celebrating success in recog...
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ARM Assembly Language Practice Question And Answer Part-1

 Hello Dear Readers, 

Today in this post I will provide some basics to advanced ARM's assembly language practice QA I have used the Keil tool for code writing.

Q-1). Perform subtraction of the numbers stored at memory location 0x4000 from the number stored at memory location 0x4004 and place the result in memory location 0x4008.

Code:

; Program of the substraction 

 area into, code, readonly 

entry 

 mov r1, #0x4000 

 ldr r2,[r1] 

 ldr r3,[r1, #4] 

sub r4,r3,r2 ; perform substraction operation 

str r4,[r1, #8] ; store the result at memory location 0x4008 

end

Output:


Q-2). Write a program to check whether the number stored at the memory location is even or odd.

Code:
; program for identity number is even or odd area prog2, code, readonly 
entry 
mov r1,#0x4000 ; data location 
ldr r2,[r1] 
tst r2,#1 ; anding with 0x01 will be updating status of flag register 
bne odd 
mov r1,r2 ; even number stored at r1 
odd mov r3,r2 ; store odd number 
end 

Output:


Q-3). Write a program to perform sum of two 64-bit numbers. Use DCD for the same.

Code:

; program of adding two 64 bit data 

area sum, code, readonly 

entry 

ldr r0,=data1 

ldr r1,[r0] 

ldr r2,[r0,#4] 

ldr r0,=data2 

ldr r3,[r0] 

ldr r4,[r0,#4] 

adds r6,r2,r4 

adc r5,r1,r3 

ldr r0,=result 

str r5,[r0]

str r6,[r0,#4] 

data1 dcd 0x4000,0x5000 

data2 dcd 0x3400,0x2000 

result dcd 0 

end

Output:


Q-4). Store a number 0xBDA35D12 at location 0x4064. Use DCD for the same.

Code:

; store given data in memory 0x4064 

area store, code, readonly 

entry 

mov r1, #0x4000 

ldr r2,=data 

ldr r3,[r2] 

str r3, [r1, #100] 

data dcd 0xBDA35D12 

end

Output:


Q-5). Swap the numbers stored at memory locations 0x4000 and 0x4004.

Code:

; program to swaping contents of the memory location without taking temporary register 

area swap, code, readonly 

entry 

mov r1, #0x4000 

ldr r2,[r1] 

ldr r3,[r1,#4] 

eor r2,r2,r3 

eor r3,r3,r2 

eor r2,r2,r3 

str r2,[r1] 

str r3,[r1,#4] 

stop b stop 

end

Output:


Q-6). Write a program to perform the below operations on 0x2020zzFF stored at location 0x4000. Perform the following operations on the number. Store result at memory location 0x4004. a. Selective-clear (on 1st byte) b. Selective-set (on 4th bit) c. Selective-complement (on 2 nd and 3rd byte) d. Selective-set (on 4th byte).

Code:

; program for the desire operation performed on the given data 

area selective, code, readonly 

entry 

mov r1, #0x4000 

ldr r2,[r1] 

and r2,r2, #0xffffff00 

orr r2,r2, #0x00000008 

mov r3,#0x0ff00000 

mov r4,#0x000f0000 

orr r3,r3,r4 

eor r2,r2,r3 

orr r2,r2, #0xff000000 

str r2,[r1, #4] 

end 

Output:




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Comments

  1. Anonymous25 January 2022 at 19:21

    Now ARM assembly language problem is also solved. Thanks for your effort.

    ReplyDelete
  2. Anonymous26 January 2022 at 06:07

    Wow bro how many parts are you uploading ?

    ReplyDelete
  3. Anonymous27 January 2022 at 23:55

    Nice posts and fabulous examples are there.

    ReplyDelete

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