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RTL Design Engineer at Skyroot Aerospace

Hello, Dear Readers, Skyroot Aerospace has a vacancy for the RTL Design Engineer role. About Skyroot Aerospace: A cutting-edge startup founded by ex-ISRO scientists. Dedicated to affordable space access, we're rewriting aerospace technology rules. Our dynamic team fosters inventiveness, collaboration, and relentless excellence. Join us on a transformative journey to redefine space possibilities. Welcome to the forefront of space innovation with Skyroot Aerospace! Purpose of role: Understand architectural requirements and Design micro-architecture, implement design blocks using VHDL/Verilog for FPGA based Avionics packages for orbital launch vehicles and ground infrastructure. Job Requirements: 2+ Years of RTL and system design experience. Strong knowledge on Digital System Design (DSD). Strong knowledge of RTL/SoC design/integration with VHDL/Verilog. Strong knowledge in problem solving and debugging skills. Ability to understand architectural requirements and Design micro-archite...
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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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