A programmer is asked to translate the following HLL code into MIPS Assembly language int isa() = (10,20,30,-5,-15,40,50,-88,19,-33); string hud="***": for (int k=0;k<10;k++) isa(k)=4*isa(k); for (int k=0;k<10;k++) cout << isa(k) << hud; // print value return 0; The programmer asked you to help him complete the missing 10 statements in the MIPS code. In your answers, please separate the opcode and the operands by ONE space only as follows: sub St9,st8,St5 .data isa: word 10,20,30,-5,-15,40,50,-88,19,-33 hud: asciiz "***" .text -globl bonus bonus: Blank 1 li $t1,10 next: Iw St9,0(St0) Blank 2 Blank 3 addi St0, St0,4 Blank 4 Blank 5 la $t0, isa Blank 6 Go: Iw $a0,0(St0) Blank 7 syscall la sa0,hud li sv0,4 syscall Blank 8 Blank 9 Blank 10 li $v0,10 # exit program syscall

A programmer is asked to translate the following HLL code into MIPS Assembly language int isa() = (10,20,30,-5,-15,40,50,-88,19,-33); string hud="***": for (int k=0;k<10;k++) isa(k)=4*isa(k); for (int k=0;k<10;k++) cout << isa(k) << hud; // print value return 0; The programmer asked you to help him complete the missing 10 statements in the MIPS code. In your answers, please separate the opcode and the operands by ONE space only as follows: sub St9,st8,St5 .data isa: word 10,20,30,-5,-15,40,50,-88,19,-33 hud: asciiz "***" .text -globl bonus bonus: Blank 1 li $t1,10 next: Iw St9,0(St0) Blank 2 Blank 3 addi St0, St0,4 Blank 4 Blank 5 la $t0, isa Blank 6 Go: Iw $a0,0(St0) Blank 7 syscall la sa0,hud li sv0,4 syscall Blank 8 Blank 9 Blank 10 li $v0,10 # exit program syscall

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Question

A programmer is asked to translate the following HLL code into MIPS Assembly language
int isa()= (10,20,30,-5,-15,40,50,-88,19,-33);
string hud="***";
for (int k=0;k<10;k++)
isa(k)=4*isa(k);
for (int k=0;k<10;k++)
cout << isa(k) << hud; // print value
return 0;
The programmer asked you to help him complete the missing 10 statements in the MIPS code. In your answers, please separate the
opcode and the operands by ONE space only as follows: sub $t9,st8,St5
.data
isa: word 10,20,30,-5,-15,40,50,-88,19,-33
hud: asciiz "***"
.text
globl bonus
bonus:
Blank 1
li $t1,10
next: Iw St9,0(St0)
Blank 2
Blank 3
addi St0,St0,4
Blank 4
Blank 5
la sto, isa
Blank 6
Go: Iw $a0,0(St0)
Blank 7
syscall
la sa0,hud
li sv0,4
syscall
Blank 8
Blank 9
Blank 10
li $v0,10 # exit program
syscall

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