Question: Please help me solve this using the software: Visual ARM Emulator ECEN 375: Computer Architecture Lab 3: Assembly: Functions Procedure: The objective of this lab

 Please help me solve this using the software: Visual ARM Emulator

Please help me solve this using the software: Visual ARM Emulator

ECEN 375: Computer Architecture Lab 3: Assembly: Functions Procedure: The objective of this lab is to understand and demonstrate how to program functions in the ARM Assembly Language. Write an assembly language program that executes the following C code: void main() { int a, b, c, d, e, f, g, h; - diffofsums (2, 3, 4, 5); b- diffofcums (3, 4, 5, 2); C = diffofsums (4, 5, 2, 3); d - diffofsums (5, 2, 3, 4); e-diffofcums(a, , c, d); ant diffofcune(int . int . inth, int i) { int result; result - diff(sunf,g), sum h, i)); return result; int sun(int , int ) { int result; result = a + b; return result; int diff(int n, int b) { int result; result - - b; return recult; Submission: Submit the Assembly file and a PDF file which suxifies and includes in screenshot of the output of , b, d, and to the Lab Submission section of Blackboard by Monday, February 17, 2020 at 1:00 PM ECEN 375: Computer Architecture Lab 3: Assembly: Functions Procedure: The objective of this lab is to understand and demonstrate how to program functions in the ARM Assembly Language. Write an assembly language program that executes the following C code: void main() { int a, b, c, d, e, f, g, h; - diffofsums (2, 3, 4, 5); b- diffofcums (3, 4, 5, 2); C = diffofsums (4, 5, 2, 3); d - diffofsums (5, 2, 3, 4); e-diffofcums(a, , c, d); ant diffofcune(int . int . inth, int i) { int result; result - diff(sunf,g), sum h, i)); return result; int sun(int , int ) { int result; result = a + b; return result; int diff(int n, int b) { int result; result - - b; return recult; Submission: Submit the Assembly file and a PDF file which suxifies and includes in screenshot of the output of , b, d, and to the Lab Submission section of Blackboard by Monday, February 17, 2020 at 1:00 PM

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