0 like 0 dislike
9.5k 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
| 9.5k 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:172.70.100.72
©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
| 6.1k 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
| 2.9k 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
| 5.3k 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
| 2.6k views
12,783 questions
183,442 answers
172,219 comments
4,824 users