0 like 0 dislike
10.6k 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
| 10.6k 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:172.70.126.16
©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
| 6.8k 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
| 3.4k 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
| 5.8k 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
| 2.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users