Why isn't that code sorting an array?

Question

#include <stdio.h>
#include <stdlib.h>

int main()
{
int x[5]={5,8,9,4};

int m=0;
int n=1;
int i =0;
int h;
for (m=0 ;m<5;m++){
for(n=1;n<5;n++){
if (x[m]>x[n]){
h=x[m];
x[m]=x[n];
x[n]=h;

}
}
}

printf("The numbers arranged in ascending order are given below \n");
for (i = 0; i < 5; i++)
printf("%d\n",x[m]);
}

Answer 1

#include <stdio.h>
#include <stdlib.h>

int main()
{
int x[4]={5,8,9,4}; // edited

int m=0;
int n=1;
int i =0;
int length = sizeof(x)/sizeof(int); // edited
int h;
for (m=0 ;m<length-1;m++){ // edited
for(n=m + 1;n<length;n++){ // edited
if (x[m]>x[n]){
h=x[m];
x[m]=x[n];
x[n]=h;

}
}
}

printf("The numbers arranged in ascending order are given below \n");
for (i = 0; i < length; i++) // edited
printf("%d\n",x[i]); // edited
}

I have modified/added 6, 11, 13, 14, 25, 26 lines.

Answer 2

Steppenwolf

good edit. Just add 1 to n = m in the second for loop's init-statement to avoid redundant iteration over container.

for (n = m+1; ...

Answer 3

Retag Tarek another version:

#include <stdio.h>
#include <stdlib.h>

int main()
{
int x[5]={5,8,9,4};

int m=0;
int n=1;
int i =0;
int h;
for (m=0 ;m<5;m++){
for(n=0;n<4;n++){
if (x[n]>x[n+1]){
h=x[n];
x[n]=x[n+1];
x[n+1]=h;

}
}
}

printf("The numbers arranged in ascending order are given below \n");
for (i = 0; i < 5; i++)
printf("%d\n",x[i]);
}

I have edited printf line with x[i ] and entire second for loop

Answer 4

C++ Soldier (Babak) the holy truth! Thank you! 🙂

Answer 5

Time Complexity: O(n²)
Space Complexity:O(1)