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About bubble sorting in C

This is my code:

//bubble sort

#include<stdio.h>
int main(){
    int a[11];
    int temp;
    int i,j;
//input

for(i=1;i<=10;i++){
    scanf("%d",&a[i]);
}

//sort

for(i=1;i<10;i++){  //the number of number
    for(j=1;j<10-i;j++) //--
        if(a[j]<a[j+i]){
            temp=a[j];
            a[j]=a[j+1];
            a[j+1]=temp;
        }
}

//output
for(i=1;i<=10;i++){
    printf("%d ",a[i]);
}

getchar();
getchar();
return 0;

}

The result was not what I expected. The input I used is 1 2 3 4 5 6 7 8 9 0, but after sorting the output is 8 6 5 4 3 7 2 9 1 0.

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2 answers


Two things:

Typo (I believe):

if(a[j]<a[j+i]){

should be

if(a[j]<a[j+1]){

Secondly,



 for(j=1;j<10-i;j++)

should be

 for(j=1;j<10-i+1;j++)

You need to run the loop over the extra time to accommodate all swaps.

Perfect link

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Note that arrays are null based - you have to run i = 0

on all instances - you only seem to need 10 numbers, so each iteration should be from 0..9

, and an array int a[10];

will suffice.

Assuming you want the bubblesort to be ordered from lowest to highest, the algorithm should be:



for(i=0; i < 10; i++){
    for(j=0; j < 10 - i - 1; j++) // Because we are looking ahead one cell, -1
        if(a[j] > a[j+1]){ // next cell
            temp=a[j];
            a[j]=a[j+1];
            a[j+1]=temp;
        }
}

The value, after the first pass, the highest number will be in a[9]

. The next cycle will set the next high at a[8]

. and etc.

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