0 like 0 dislike
8.3k views

Please write a program that receives 5 numbers in an array, then makes those numbers in ascending order by using bubble-sort.

**Use the template given below**

Code (Template)

#include <stdio.h>

int main(){
	int number[5]; // Do not change
	int size = 5, temporary; // Do not change
	
	for(int i=0;i<size;i++){ // Do not change
		scanf("%d",&number[i]); // Do not change
	} // Do not change

	// Your conditions start from here

	return 0;
}

 

Input 1

77 22 88 11 22

Output 1

11 22 22 77 88

 

Input 2

34 32 12 64 2

Output 2

2 12 32 34 64
[Exercise] Coding (C) - asked in Chapter 4: Arrays by (5.9k points)
ID: 37580 - Available when: Unlimited - Due to: Unlimited
| 8.3k views

46 Answers

0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
** ** * * *** ** ** * ** x[] = {77,22,88,11,22};
*** ** * ** * *** ** * size = 5;
    
* * ** ** *** *** print the array */
** ***** *** *** * * * i = 0; i < size; i++){
* ** * *** ** ** ******** * * **** **** *** * * * * ", x[i]); ** *** ** * * * * ** *******
** *** ** * * **** *
* * * * * * * * ** * * * ********
**** ** * ** * **
* ** **** ** * I need this to happen size-times */
* * * ** * * **** ** ** * j = 0; j < size; j++){
* ** *** **** * ** * * ** * * * *** *** from left to right */
** * *** * * * ** * *** * * **** *** * * ** i = 0; i < size-1; i++){
* ********* * * ** * * **** * * ** * ***** *** * * ** ** * *** **** * * i-th element with its neighbor to check if they are in order */
** * * *** ***** * ** * * * * **** * * * * * **** * * ** * ** > x[i+1]){
** * ** * ** ***** ** ** * * ** * * ** * *** ** * * *** ****** * ** * * ** * * * ** * exchange the values of x[i] and x[i+1] */
* * ** * ** * * **** *** *** ** * *** * ***** *** * *** ** ** ***** * *** *** * *** * temporary; ******* * * **** * * * * * * *
** * * * ** **** **** ** ** * **** ***** * **** * * * ***** **** ** * ** * ** * ** ** = x[i];
* ** * * ***** * **** * **** ** * * * * * * ** * * * ** * ** ***** ** * * = x[i+1];
* * * * * * * * **** * ***** * **** *** *** * *** * ** * * ** **** * * * * * * ** *** = temporary;
* * ** ** * *** ***** * * * * ** ***** * ** *** ***** ** * *** * * *** *
** * *** * *** * ** ** * ******* * * ** * ***
* * * ** ***** * *
** **** *** * * * *
* * **** * ** *** ** print the array */
* * ** * *** ** * *** *** * * ****** * sorting: \n");
** * * * * * * ** ** i = 0; i < size; i++){
* *** * * * ** *** ** *** * **** ** ** ** *** **** ", x[i]); ** * ***** *** ** * ***
*** ** * * *** ** *
* ** * **** **** **** *** * * * *
* * * **** * *
* * * ** * *** **** *
*** * ** ** ******* * * ** * *
*** * ** * ** * * * ** ** * ** * ** * * ***  
* * * ***** * * ** ** 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
** **** * *** * * ** x[] = {77,22,88,11,22};
*** ** * * * * size = 5;
    
* * **** * ** * *** print the array */
* ** *** ** * * *** i = 0; i < size; i++){
* ** ***** ** * **** **** ***** * **** * *** * * * ** ", x[i]); ** ** ** * * * *** *** * *
** * * ***** * * * *
* ** **** ** **** **** *** * ** * *
*** ***** * * * * **
******* *** ** ** * * * I need this to happen size-times */
* *** * * ** **** **** j = 0; j < size; j++){
* * ******* * * *** * * * **** *** ** * *** ** from left to right */
* ** * * * **** ***** * ** * *** ** **** ** ** * i = 0; i < size-1; i++){
*** * * *** *** * * **** * *** * * **** *** * ** ** **** * *** * i-th element with its neighbor to check if they are in order */
** * * ****** * * ***** * ** * ** * *** * *** ** ** ******* > x[i+1]){
*** *** *** **** *** * * * ***** * ********* ** * ********* *** ** ***** * * ***** exchange the values of x[i] and x[i+1] */
* * *** *** * * * * **** * * * * * *** * **** ** **** * *********** **** * ** **** temporary; ****** * *** ********* * *
* ** ** ***** ***** * * * ** ** * * ** * * ** * * * ** *** *** *** * * ** ********* ** * *** ** = x[i];
* ** * ******* *** * ** * * ********* ** ** *** *** * **** *** * ***** * *** ** ** = x[i+1];
*** *** ****** *** *** *** * * * ** ** *** **** ** * ***** *** * * * * *** ** **** * = temporary;
** * * * * * ***** ** * ** ** ******* * ** * * * ** ** * * *
* ** ** * * * ***** * **** * ** * * ** ** * **
* * * ** * * *** *
* * **** *** * * ** * *
*** *** * * ** print the array */
*** * ** * * ** * * * ** * * sorting: \n");
*** * *** *** * * * i = 0; i < size; i++){
* *** * ******* * * * * ****** ** * * * * * * ** * **** ** ", x[i]); * ** * * * *** ** * * * ****
* ** *** ** ** *
**** * * * * ** * ** ** *** * *
*** * ** * **** **
* ** ** ** * *** *
* ** * * * * * * **** * * * * * * * *
*** ****** *** * ** ** ** * ** **** *** * * **  
* * ** * * * * * 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
******** * * * * * * ** x[] = {77,22,88,11,22};
* * ** * * * ** * * size = 5;
    
*** * * * * * * * ** ** print the array */
**** ** ** * * * * i = 0; i < size; i++){
* *** * * ****** * **** * ** ** * ** * * * *** * * ", x[i]); * * ** ** *** * *** *****
* * * * ** ********* **
* * ****** ** * ** * * * ** * ** *** *
*** * * * ** * * ** **
******* * * * * * * *** I need this to happen size-times */
** * * * ** ****** *** * * * j = 0; j < size; j++){
* ** *** ** *** ******* ** * *** * ** * * * from left to right */
** * * * ** ******* * * * ** * * * * ** i = 0; i < size-1; i++){
* * ****** * * * ** **** ** *** * ** **** * *** * * ** * * * * * ** i-th element with its neighbor to check if they are in order */
* * * * * * *** * *** *** **** * ******** ** * ** ** * * * ** *** ** > x[i+1]){
* **** * **** *** * * * ** ** ***** * *** * **** * * * ** * ** * ** ** * ** exchange the values of x[i] and x[i+1] */
****** ** *** * * ** ** **** * *** ** * * * * * * **** * * ** * ***** ** * * * * * temporary; * * * * * ********** **
***** ** * * ** ** * ** * * ** ** **** ** ** * ****** ** * *** ** ** * * ** *** *** ** * * *** ** = x[i];
** * * *** ****** * * *** *** *** ** ** **** * * * * **** ** ** * ** * ******* ** ** ** * * = x[i+1];
** **** * * *** * * * * * ****** *** * ***** **** ** *** ** * * * ** *** ** * ** * ** ** = temporary;
* ** * * *** * ***** *** ** * **** ** * *** ***** * ** ** * **
* * ** **** * **** * ** * * * * * **********
*** * * * **** ** **
**** * * *** * ***** * ** *
*** *** *** ** ** * * *** * print the array */
** * **** ***** ** * * *** * * * * * * sorting: \n");
*** * ** *** * * * * * i = 0; i < size; i++){
** *** * ** ** * * * *** * * * * * * ** ** * ** ", x[i]); * ** ***** * ** ** ** ** * ** *
* ** **** ***** ** *
** ** ** ** * ** * * * * **** * ** * * **
** *** * *** **** ** *
**** *** ** ** * *** * ***
*** * *** * * * * * ** ** * *****
* * ** * **** **** * ** * * ** **** ** ** *  
* * * *** * * ****** 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
* ** ** *** * ** * * * x[] = {77,22,88,11,22};
* * ** * * *********** size = 5;
    
* * * * **** * ****** print the array */
* * * ** **** *** * * ** i = 0; i < size; i++){
* * **** *** * ** * ** * ** ** * **** *** * * ", x[i]); * * * **** * * * * ** * ***** **
* * ** *** * * *** * **
* ** * * ** * ** * * * **
** **** *** * * ***
* * **** ****** ** I need this to happen size-times */
*** ** * * * ** * ** * j = 0; j < size; j++){
**** * ** ** ** *** * ****** * * ** ** * * from left to right */
**** * * * *** * *** * * * * *** * *** ** * i = 0; i < size-1; i++){
** * ** * ** **** ** * **** * ** * ** ** * * * **** * ** *** * ***** i-th element with its neighbor to check if they are in order */
***** ** * ** *** **** * * ** ***** *** * *** ** *** * * * * > x[i+1]){
** ********* ** ** ******* * ** * ***** ** ** ***** * ** * ** ** *** ** * ***** * exchange the values of x[i] and x[i+1] */
*** * ** * * * * *** * * * * * * * * * * * ****** ** * ** * ** * * * temporary; ** * ***** *** * * * *** *
* * * ** * * *** * ** ** ** *** *** ** * * *** * * ** * * ** *** ** ** * ** *** * * = x[i];
* *** * * * ** ** * * ** ** ** ** * *** ** ** * **** * *** ** * * * * *** **** = x[i+1];
* * * *** **** ** *** ** * * * * * ****** ***** * * **** * * * * *** * ** ** * * = temporary;
* ** * * * * *** **** ***** ** * ** **** * **** **** ** ** * * **
***** ** ** ******** * ** * * ** * *****
* * **** * ** * *
* ** ***** * * ** * **
** * **** ** * print the array */
* **** * * ** ** ** *** **** *** sorting: \n");
* ** * ***** ** ** *** i = 0; i < size; i++){
* ** ** * *** *** * ***** ** * ** * ***** * ** * ** ", x[i]); * * * ** *** **** ** ** * *
*** * * * * *
** * * * ** * * * ***** * * * * * *
* ** ** **** * * ****** *
* * *** **** ** * ***
* * ** ** ** * ** **** * ** *
** ** ***** ** * * ** *** * * * * * * * **** *  
** * * *** *** ** ** * * 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
****** * ** * ** * * x[] = {77,22,88,11,22};
* ** * * * * * * * * size = 5;
    
* * * ** ** * ** print the array */
* * * * *** * * * * ** * * i = 0; i < size; i++){
** ** * *** * * ** * ** * * *** * ** * * * * ** **** ", x[i]); ** * **** ***** ********* ** *
** *** ** ** * * *
** * ** * *** * *** ** * * * *
* ***** ******** ** **
** ** ** * * **** *** I need this to happen size-times */
* ** ** ** **** ** * *** * * j = 0; j < size; j++){
* **** * * ** * * * ** *** * * * * * * from left to right */
**** *** *** * ** * * ******** * ***** * * * ** * i = 0; i < size-1; i++){
* * *** * * * ** *** *** ** * ** * * * ******* * * * *** i-th element with its neighbor to check if they are in order */
* ** ** *** *** ** * * ** *** * ** * ** * *********** ** *** * * * ***** > x[i+1]){
* ******* *** ** ***** *** * * * * *** ** ** *** ***** ** ** ** * * ** *** *** ** exchange the values of x[i] and x[i+1] */
** * * * * **** * * **** * ***** **** * ** * * *** ** ** *** * *** ** * * **** * * ***** ** * * temporary; **** ** * *** ** * ****** * *
* * ** *** ** * * * * ********** **** ** * *** * ****** *** *** * * *** *** * ** ** * = x[i];
******* * * ****** * * * * * * *** * **** * **** *** **** * * ** *** ** *** * * = x[i+1];
* * * ** ****** * * * * * ** * * **** * *** ** ** **** ** * **** * * * * * = temporary;
** ** * ** ** *** * *** ** **** * * * * ** * ** * **** ** * * *
* * * ** * * * **** ** ** * ****** ** * * * *
* ** * **** *** **
* * ** * * **** * **** *
**** ** * ** * ****** * * print the array */
* * * **** * ** * * ***** * *** sorting: \n");
* ** * *** * * * ******** i = 0; i < size; i++){
*** ** * **** ** * * * * * ** * ** * * * * * ", x[i]); ** * * * * ** * * *
* ** *** ** **** * ***
** * * * * **** * ** ** ** *** * **
* * * *** * **** * ****
* *** *** * ** *
* * ****** * ** * ** ** ** **
* *** ** ** * ** *** * * * *** *** * * ** *  
***** * * * * * ** 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
* * *** * ** ** ** x[] = {77,22,88,11,22};
** * *** * ***** * * size = 5;
    
* * * ******** *** print the array */
** * * * * ** * * ** ** i = 0; i < size; i++){
*** * * * * **** ** ** * **** ****** * * ** ****** ** **** ", x[i]); ** * * * **** ** ***** * * ** * **
* ** * ** ** *** *** * *
** * * * * * * ** ** * * * * ** ***
* *** *** ** ** ****
** * ** * * ******* I need this to happen size-times */
* * ** *** * ** ***** * j = 0; j < size; j++){
**** * ** * * *** *** * * * * ** * *** from left to right */
*** * * * ** * *** *** **** * ** *** i = 0; i < size-1; i++){
**** * ** * *** * * *** *** * ** * * * * * ** * ** * * * * i-th element with its neighbor to check if they are in order */
* **** * ** ******* * ** * * * ** ** * * ***** ** * **** * > x[i+1]){
** * ** *** *** ** ** * * * * *** *** * * ** ** ******* * ** * * ** * ** * *** * exchange the values of x[i] and x[i+1] */
* ** ** * *** * ** ** * *** ** ** ** ** * *** **** ** * *** ***** * *** * * temporary; ** **** ** * * * * * ** ******
** *** * * * * * * **** * ****** ** * * *** * *** ** *** * * * * * * ** ** * **** ** * * * * = x[i];
* ** ** * *** * * **** *** * ** * * * **** * * * * * ** *** * * * ** ** * ** * * * * * * = x[i+1];
* *** ** ** * * ** ** * ** * ** ***** * *** ** * *** * * *** **** *** * ** * * *** *** **** ** *** = temporary;
** * * * * ** * ** * * **** *** ** **** ** **** ******* ** * *
* **** * *** ** *** ** **** * **** *** * * **
****** * * **** *
* * * ** ****
* ** *** * * * * * *** *** print the array */
**** ** *** *** * * *** * * ** * * sorting: \n");
* * * * *** *** * * ** i = 0; i < size; i++){
** * * * *** * * ** *** * ** ** * * *** * * ", x[i]); * * * * * *** ** *** *** ** *
* * ***** **** *****
** * *** ** *** ** ** ****** * **** * *
* * * ** *
* ** * * *
** * ***** ** *** ** ******** *** ***** ** ** *
***** ** ** * * * * * * ** * * **** * * ***  
* ** * * ****** ** 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
* * **** **** ** * * * x[] = {77,22,88,11,22};
* ** * * ** ***** * size = 5;
    
* ** **** ** ** * ** print the array */
* * ** ** *** ** * i = 0; i < size; i++){
* ** * ** ** ** *** **** * ** ** ** * ** * * * ", x[i]); **** * * *** ** ** *** *
* ** * ** *** *
* * ** ** **** ** **** * *** ** *** *
*** ** * * * ** * *
*** **** **** ***** I need this to happen size-times */
*** * **** * * *** j = 0; j < size; j++){
**** ** * * * * * *** * * ** ** ** *** * ** ** from left to right */
** * * * ***** ** ** ** ***** * **** * **** * * * * i = 0; i < size-1; i++){
* * * * * ** *** *** * *** * ****** * **** *** * * * ***** ** i-th element with its neighbor to check if they are in order */
***** **** ** * *** * * * ** * *** * *** ** * * * * * * ** * > x[i+1]){
***** ** *** ** * ** *** * * * ** ** * * ** * * *** ** ** *** ** * * ** ** *** *** * * * exchange the values of x[i] and x[i+1] */
** ** * * **** ** * *** * ** ** *** * ******* * ** ** ****** * * **** * ** * * * temporary; ** **** *** ** * *** * ** **
** * * ** * * ** ***** ** ** ** *** * * * ** ** * * ***** *** **** * * * ** * = x[i];
*** * **** * *** * * **** * * * ** ** * * ** * *** *** * ** * *** * ** * * * = x[i+1];
** ** **** * * * ** ** * * * *** * ****** ** * * * * **** ** * ** * * * ** *** * * ** = temporary;
* * * ** * * * ** ** * *** ***** ** * *** * * *** * * * *
** ** * * *** * ** * ** * * ** *** *
*** * ** * * * *
*** ****** * ** * ***
** * * *** ** ** ** ** print the array */
** ** * * * ******* *** * **** sorting: \n");
* * ** * * * * i = 0; i < size; i++){
** * **** *** ** * ** ** * ** * * * ** * ", x[i]); ** ** ** * * * ** * * * ** *** **
* * * * * * *** **
* *** * ****** ** * ** * ** * *** * * * *
*** * ******* * * * * * *
** **** * * * **
** ***** ** *** *** * ** * * ** *** ** *
* * * * * *** * *** *** * ** **** * * ***  
***** * * **** ** ** 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
*** **** * * *********** *** x[] = {77,22,88,11,22};
***** * ** * * * * size = 5;
    
* ** * * * * ** * print the array */
*** ** ** * * ** * *** * i = 0; i < size; i++){
** * ** ****** *** *** * * * * *** * *** * * ", x[i]); * * ** ** *** * * **** ** **** **
* ** ** ** *** * **
** ** * * *** ** *** * *
** ** * *** **** * * *
* **** * *** * * * * I need this to happen size-times */
** * *** * *** * *** ** j = 0; j < size; j++){
** *** * *** ** *** **** ** ** * * * **** * * from left to right */
* *** * ** *** ** * * * ** ** * * * *** ** * i = 0; i < size-1; i++){
*** ** * *** * * ** ** *** ** ** * * * **** *** *** *** **** *** * i-th element with its neighbor to check if they are in order */
** * ** * * * ** *** * * * * ** *** * * * * * *** * * > x[i+1]){
****** ***** ** * * * ** ** * * * * *** * ***** ** ** * ** * * ******** * *** * *** * exchange the values of x[i] and x[i+1] */
** **** * * **** **** * * * * * * ** ** *** * ** * ***** ** *** ** ** * * * * * ** * * * ** temporary; * ** ** ** * *** * *
* * * ***** ***** ** ** * **** * * ** * ** ** ** ** * ** * *** * * ** * ***** ** ** ***** *** = x[i];
* ** * ** *** * **** *** *** ** ** * * * ** *** ***** * * *** ***** * * * * * * * ** * ** = x[i+1];
* * * * * * **** * ** * *** ***** * * * ** * * * * **** * ** ** * * * * ** * * = temporary;
* ** * ** * *** ** * * * * **** * ** * * **** **** **
**** *** ** **** * * ** * *** *** *** *
* * **** ***** **** * *
** * * ** * *** ***
* ** ** * *** ** *** ** print the array */
*** ** ****** * * * * ***** * sorting: \n");
* *** *** * * * i = 0; i < size; i++){
* * *** * * *** ** * * ** ** ** * ** * * * * * ** ", x[i]); * * *** ***** ** * ****** ** ** * **
*** ** **** * * *** *** *
****** ** * ** * * ** * * *** * * * ** *
** ** ** *** **
* ** ** ** *** *** *
** *** ****** * * * ***** * ** * *
** ** ** * ** * ** ** * ******** **** *  
** ** ** **** *** * 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
** * ** * * *** ******* x[] = {77,22,88,11,22};
*** * * ***** ** * ** size = 5;
    
* ** ** * * * **** print the array */
* ** ** * *** * i = 0; i < size; i++){
* ****** ** ***** ** * * ** ** * **** * ** ** ** * ** ** * ", x[i]); * **** ** ** * * * * ** * * * * *
* * ** * * * ** * ** **
* **** * ** * ** * ** * ****
** * * ** * ** * ** * ****
* *** ** ** * * ** *** * I need this to happen size-times */
* * * * * * * ** * ***** j = 0; j < size; j++){
* *** ** ** * * ** *** ** ** ** * ** * from left to right */
*** * * ** ** * **** **** ** ** * ***** ** * ** i = 0; i < size-1; i++){
* *** * *** * * ** ** ** * * ** ** ** ****** * ** **** **** * ****** * i-th element with its neighbor to check if they are in order */
*** * * ** ** *** * ** * * * * * ***** * * * * ** ** * * ** * *** * **** * * > x[i+1]){
** * * * ** ** ***** * * * ** * *** ** ** ** * *** * * ** ** * * * ** ** ** ** * * exchange the values of x[i] and x[i+1] */
** * * ** ***** * * * * ***** **** ** **** *** **** * ** * *** *** * *** * temporary; *** * ***** * * * * ** * * ****** *
*** * ** ****** * * * ** *** * * * * *** ****** *** ***** ** ** * * *** * **** *** ** ** * * ****** = x[i];
*** *** **** ** * * * * **** * * **** *** * ** * ***** **** **** *** ** ** ** * ** * ***** = x[i+1];
**** * ** *** *** ** ** * ** * * * * ** *** ** * ** ** * ** ** * * * * * = temporary;
* **** * * *** * * * * ** ** * * * ** * * ** *** * *
********* * * * *** * * *** *** ** ****** * *** **
** * * * ****
* ********** ** * * * ** **
* * *** * * * ******* * ** print the array */
** **** ** ** *** **** sorting: \n");
* ** * * * ** * ** * ** i = 0; i < size; i++){
* **** * *** * *** * ** ** ** ** *** * * * * ** ** **** * *** ", x[i]); * * * ** ** **** ** **********
************ * * ******
* * ******* * ** ** **** *** * ** * *
* *** * *** * *
* ** * ** * * * * *** * *
***** * ** * * * **** **** ** *
**** * * ******** * * ** ** * ** ** ***  
** * *** * * *** * ** * * 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
* * ** ***** * * ** * x[] = {77,22,88,11,22};
* * ** * * * size = 5;
    
*** * **** * ** print the array */
******* *** * ** * * * * i = 0; i < size; i++){
* * * ***** ** * ** **** ** * ***** ** ******* ", x[i]); ****** * * *** * ** * ***
* ** * *** * ** * *
** *** * * ** * * **** **** * ***** * **
** * * *** * ** **
* * * * * ** * *** * * I need this to happen size-times */
* ** * * *** * *** * * ****** j = 0; j < size; j++){
* *** * ** ** * * ***** ** **** * * from left to right */
**** ** ** * * *** *** **** * * *** *** * i = 0; i < size-1; i++){
* ** * *** ** *** * *** **** *** * * * * ** *** **** *** *** * i-th element with its neighbor to check if they are in order */
** ** ** * * ** ** * ** * * * ** * *** ** * ** * ** ***** * *** * *** > x[i+1]){
** *** ** ** ** ***** ***** ** * ** *** * ** * * * ** ** ** * * **** *** * * * ** exchange the values of x[i] and x[i+1] */
* ***** * * * **** ** * ** * *** *** * ********* * * * * ** ***** * ** *** *** **** ** ** * ** temporary; ** * * ** ***** ***** *
*** ****** *** *** * ** * * * ** ** * ** * ** *** ** * ** * **** ** ** * * * ** * * * * *** = x[i];
** ** ** * * * * *** * * * ** ***** * ** ***** * *** **** * ** ** ** ** * * * ** = x[i+1];
** ** * * * ** * ** * * * ** *** ** ** * * * * *** ** *** ** * * ** ******* * * = temporary;
* * ** * ** * * * ** * * * *** * ** ** **** * * * ** *** * ** *
** ***** *** * * ******* * ***** * * * * ****
* **** ** * ** * * **
* ******* ** *** * * * *
* * ******* ** ** * **** print the array */
** * ** ** ** * * * ** **** ** * * * *** * ** sorting: \n");
*** * * **** ** *** * * i = 0; i < size; i++){
* *** * * ** ** * *** * *** * * * * * * ** ** *** ", x[i]); ** ** *** *** * ** ** ** *** **
*** * ****** *** * * *
* ** * * * ** * * * * ** * *
* *** * ** * *** *
***** ** ** *** * * * * *
* ** ****** **** * * *** ** ** * * **
** * ** * * ** * **** * ****** * ****** * * * * *  
* * * **** * * * * 0;
}
answered by (153 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.71.255.35
©2016-2025

Related questions

0 like 0 dislike
0 answers
[Resource] asked Dec 15, 2017 in Chapter 4: Arrays by nat236919 (5.9k points)
ID: 37571 - Available when: Unlimited - Due to: Unlimited
| 6 views
0 like 0 dislike
18 answers
[Exercise] Coding (C) - asked Jan 5, 2018 in Chapter 4: Arrays by nat236919 (5.9k points)
ID: 41054 - Available when: Unlimited - Due to: Unlimited
| 5.1k views
0 like 0 dislike
20 answers
[Exercise] Fill in the blank - asked Dec 15, 2017 in Chapter 4: Arrays by nat236919 (5.9k points)
ID: 37578 - Available when: Unlimited - Due to: Unlimited
| 2.5k views
0 like 0 dislike
17 answers
[Exercise] Coding (C) - asked Dec 7, 2017 in Chapter 4: Arrays by nat236919 (5.9k points)
ID: 35804 - Available when: Unlimited - Due to: Unlimited
| 4.4k views
0 like 0 dislike
18 answers
[Exercise] Fill in the blank - asked Dec 7, 2017 in Chapter 4: Arrays by nat236919 (5.9k points)
ID: 35798 - Available when: Unlimited - Due to: Unlimited
| 2.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users