Computer

#include <stdio.h> #include <stdint.h> #include <stdbool.h> extern uint32_t aTask1(int32_t *arr, uint32_t n); extern uint32_t aTask2(uint32_t n); extern uint32_t aTask3(uint32_t n); extern uint32_t aTask4(uint32_t n, uint32_t b); // Global variable int32_t array_e[8] = {-1, 5, 3, -8, 10, 23, 6, 5}; uint32_t rslt1, rslt2, rslt3, rslt4; uint32_t num = 0xAAAAAAAA; // For loop: Find the index of the min number in an array arr of size n uint32_t cTask1(int32_t *arr, int32_t n) { uint32_t minLoc = 0; return minLoc; } // While loop: Determine the number of binary 1's using the parity-checking alg uint32_t cTask2(uint32_t num) { uint32_t count = 0; return count; } // Do-while loop: Count the number of binary 1's using a simplified approach uint32_t cTask3(uint32_t num) { uint32_t count = 0; return count; } // Any loop: Count the number of digits (in terms of base b) of a number uint32_t cTask4(uint32_t num, uint32_t b) { uint32_t count = 0; return count; } int main(void) { rslt1 = cTask1(array_e, 8); printf("C version of Task 1: rslt1 = %d \n", rslt1); rslt1 = 0; rslt1 = aTask1(array_e, 8); printf("ASM version of Task 1: rslt1 = %d \n", rslt1); rslt2 = cTask2(num); printf("C version of Task 2: rslt2 = %d \n", rslt2); rslt2 = 0; rslt2 = aTask2(num); printf("ASM version of Task 2: rslt2 = %d \n", rslt2); rslt3 = cTask3(num); printf("C version of Task 3: rslt3 = %d \n", rslt3); rslt3 = 0; rslt3 = aTask3(num); printf("ASM version of Task 3: rslt3 = %d \n", rslt3); num = 0123456; rslt4 = cTask4(num, 8); printf("C version of Task 4: rslt4 = %d \n", rslt4); rslt4 = 0; rslt4 = aTask4(num, 8); printf("ASM version of Task 4: rslt4 = %d \n", rslt4); while (1); }