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addi $sp, $zero, 6000   # Assume Stack memory starts at 6000. Do not modify. # Procedure Main (Do not modify) Main:      add $a0, $zero, $s0      add $a1, $zero, $s1            jal Sum            add $s2, $s0, $s1      add $s2, $s2, $v0            j End # Procedure Sum Sum:      # Type your code here.       # Procedure Dif (Do not modify) Dif:      sub $v0, $a1, $a0      jr $ra End:

addi $sp, $zero, 6000   # Assume Stack memory starts at 6000. Do not modify. # Procedure Main (Do not modify) Main:      add $a0, $zero, $s0      add $a1, $zero, $s1            jal Sum            add $s2, $s0, $s1      add $s2, $s2, $v0            j End # Procedure Sum Sum:      # Type your code here.       # Procedure Dif (Do not modify) Dif:      sub $v0, $a1, $a0      jr $ra End:

Computer Networking: A Top-Down Approach (7th Edition)
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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addi $sp, $zero, 6000   # Assume Stack memory starts at 6000. Do not modify.

# Procedure Main (Do not modify)
Main:
     add $a0, $zero, $s0
     add $a1, $zero, $s1
     
     jal Sum
     
     add $s2, $s0, $s1
     add $s2, $s2, $v0
     
     j End

# Procedure Sum
Sum:
     # Type your code here.
     

# Procedure Dif (Do not modify)
Dif:
     sub $v0, $a1, $a0
     jr $ra

End:

Given the following C program and the mapping of registers to variables, complete the MIPS implementation of procedure Sum. int Dif(int a, int b) { return b a; } int Sum(int m, int n) { int p Dif(n+1, m-1); int q Dif(m+1, n-1); return p + q; } int main() { int X, У; Z %3D х + у+ Sum (х, у); return 0; }
Transcribed Image Text:Given the following C program and the mapping of registers to variables, complete the MIPS implementation of procedure Sum. int Dif(int a, int b) { return b a; } int Sum(int m, int n) { int p Dif(n+1, m-1); int q Dif(m+1, n-1); return p + q; } int main() { int X, У; Z %3D х + у+ Sum (х, у); return 0; }
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