What is the difference between an exception and an interrupt, and what is the role of the stack in either case?
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What is the difference between an exception and an interrupt, and what is the role of the stack in either case?
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- A procedure of X86-64 can pass some arguments (e.g., up to the first six arguments) via registers rather than the stack frame. Question 2 options: True FalseWrite a sequence of statements that display a subroutine’s return address. Hint: the esp register points to the return address stored on the call stack. You can call procedure WriteHex. This can be solved in 2 lines of code. The following exercise can be solved using either 32-bit or 64-bit code. Assembly language., which contains temporary data (such as 7. A process generally also includes the process . function parameters, return addresses, and local variables), and a contains global variables. which ..... stack / data section heap / data section stack / code section heap / data section
- Write a sequence of statements that display a subroutine’s return address. Hint: the esp register points to the return address stored on the call stack. You can call the procedure WriteHex. This can be solved in 2 lines of code. write a full code in Assembly LanguageWhat is the function of software security in the stack and the development process of a certain application?MIPS Simulator QtSpim: You are to have a complete program in MIPS assembly language that behaves exactly as the included C program. This program contains four functions in addition to the main() one. Your solution must contain all five C routines as they have been coded in the example. Make sure to run the program in MIPS and show the same output on MIPS as well to make sure there are no errors. Below is the five C routines and attached is the image of what the output must print out on QtSpim. #include <stdio.h> int getMax(int arr[], int n){int mx = arr[0];for (int i = 1; i < n; i++)if (arr[i] > mx)mx = arr[i];return mx;}void countSort(int arr[], int n, int exp){int output[n];int i, count[10] = { 0 };for (i = 0; i < n; i++)count[(arr[i] / exp) % 10]++;for (i = 1; i < 10; i++)count[i] += count[i - 1];for (i = n - 1; i >= 0; i--) {output[count[(arr[i] / exp) % 10] - 1] = arr[i];count[(arr[i] / exp) % 10]--;}for (i = 0; i < n; i++)arr[i] = output[i];}void…