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- В.width; } { t = B.type; w = { T.type = C.type; T.width = C.width; } T → B C В > int { B.type = integer; B.width 4; } В — foat { B.type = float; B.width = 8; } { C.type = t; C.width = w; } C - [ num] C1 = array(num. value, C1.type); { C.type C.width = num. value x C1. width; }Match the C-function on the left to the Intel assemble function on the right. W: cmpl $4 movl %edi , %edi jmp .L4(,%rdi,8) %edi .L3: movl $17, %eax ret .15: movl $3, %eax int A ( int x , int y) { int a ; if ( x == 0 ) else i f ( x == 1 ) a = 3 ; else i f ( x == 2 ) a = 2 0 ; else i f ( x == 3 ) a = 2 ; else i f ( x == 4 ) a = 1 ; ret .L6: a = 17; movl $20, %eax ret .L7: movl $2, %eax ret else a = 0; .L8: return a ; movl $1, %eax .L2: ret . section .rodata . L4: .quad .L3 .quad .L5 .quad .L6 .quad .L7 .quad .L8 X: testl %edi, %edi je cmpl je cmpl je стр1 je cmpl .L16 $1, %edi .L17 $2, %edi .L18 $3, %edi int B (int x, int y) { int a; switch (x) { .L19 $4, %edi %al movzbl %al, %eax case 0: a = 17; break; sete break; case 1: a = 3; case 2: a = 20; break; case 3: a = 2; break; case 4: a = 1; a = 0; } return a; ret .L16: break; movl $17, %eax ret .L17: movl $3, %eax } ret .L18: movl $20, %eax ret .L19: movl ret $2, %eaxint a[ 10] = {1,5,3,8,4,2,6,9,7}; int *p; p = &a[2]; p[0] = 10; p[1] = 15; What are the values of a? Select one: O a. {10,15,3,8,4,2,6,9,7} O b. {1,15,10,8,4,2,6,9,7} O c. {1,5,10,15,4,2,6,9,7} O d. {1,10,15,8,4,2,6,9,7}
- 2 = 0 ) { numList [ i ] i } } } [1, 2, 3, 4, 5] [0, 2, 4, 6, 8] [2, 4, 6, 8, 10] [1, 3, 5, 7, 9]-0 FAMAS oro ag -0 1 void printVerticalGraph(int nUnits) { 2 4 9 10 11 12 13 14 15 16 17 456 BARHEIDEN 18 19 20 21 22 23 24 25 26 27 int nRev; printf("%d\n", nUnits); while (nUnits>0){ nRev = getReverse (nUnits); // 1) Adjust nUnits ?? [Q1] ?? while (nRev>0){ nUnits *= 10; // 2) Condition to print * if(?? [Q2]??) { printf("*"); } // 3) Update for nUnits for * case ?? [Q3] ?? }else{ printf(" "); // 4) Update for nUnits for ?? [Q4]?? } // 5) Update to nRev ?? [Q5]?? printf("\n"); case#include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…
- 5- int arr[4]= {2, 4, 5, 3); for (int i= 0; i<4; i++) cout<* A = [2-31; 20 -1; 1 4 5], then det(A) is 49 39 27 33 Other:if L1={ab,b} L2={abb,bb} then L1L2= * O {ababb,abbb,abbab,bbab} O {ab,b,abb,bb} O {ababb,abbb,babb,bbb} O {abbab,abbb,bbab,bbb}Q1 #include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…C++ A robot is initially located at position (0; 0) in a grid [?5; 5] [?5; 5]. The robot can move randomly in any of the directions: up, down, left, right. The robot can only move one step at a time. For each move, print the direction of the move and the current position of the robot. If the robot makes a circle, which means it moves back to the original place, print "Back to the origin!" to the console and stop the program. If it reaches the boundary of the grid, print \Hit the boundary!" to the console and stop the program. A successful run of your code may look like:Down (0,-1)Down (0,-2)Up (0,-1)Left (-1,-1)Left (-2,-1)Up (-2,0)Left (-3,0)Left (-4,0)Left (-5,0)Hit the boundary! or Left (-1,0)Down (-1,-1)Right (0,-1)Up (0,0)Back to the origin! About: This program is to give you practice using the control ow, the random number generator, and output formatting. You may use <iomanip> to format your output. You may NOT use #include "stdafx.h".#ifndef lab5ExF_h #define lab5ExF_h typedef struct point { char label[10]; double x ; // x coordinate for point in a Cartesian coordinate system double y; // y coordinate for point in a Cartesian coordinate system double z; // z coordinate for point in a Cartesian coordinate system }Point; void reverse (Point *a, int n); /* REQUIRES: Elements a[0] ... a[n-2], a[n-1] exists. * PROMISES: places the existing Point objects in array a, in reverse order. * The new a[0] value is the old a[n-1] value, the new a[1] is the * old a[n-2], etc. */ int search(const Point* struct_array, const char* target, int n); /* REQUIRES: Elements struct-array[0] ... struct_array[n-2], struct_array[n-1] * exists. target points to string to be searched for. * PROMISES: returns the index of the element in the array that contains an * instance of point with a matching label. Otherwise, if there is * no point in the array that its label matches the target-label, * it should return -1. * If there are more than…SEE MORE QUESTIONS