Using insertion sort c++

Question

I wrote a code for insertion sort and there appears to be no errors (it compiles fine), but it doesn't print anything or ask for a user input. I have looked over this several times and I can't figure out why the code won't run properly. Thanks!

#include <iostream>
using namespace std;

void getInput(int a[ ], int n);
void insertionSort(int a[ ], int n);
void print(int a[ ], int n);

int main()
{
    int n=7;
    int a[n];

    getInput(a, n);
    insertionSort(a, n);
    print(a, n);

    system("pause");
    return 0;
}


void getInput(int a[ ], int n)
{
    for(int i; i<n;i++)
    {
        cout<<"Number? ";
        cin>>a[i];
    }
}

void insertionSort(int a[ ], int n)
{
    int temp, j;
    for(int i = 0; i<n; i++)
    {
        temp = a[i];
        j=i;

        while(j>0 && a[j-1] > temp)
        {
            a[j]= a[j-1];
            j=j-1;
        }
    }
}


void print(int a[ ], int n)
{
    for(int i= 0; i<n; i++)
    {
        cout<<a[i]<<"    ";   
    }

    cout<<endl;
}

Hello, welcome to StackOverflow! Have you trying debugging through your code step-by-step to see where the issue is? — John Odom
– John Odom, Commented Jan 21, 2015 at 19:51
@JohnOdom Yeah, and it compiles fine it's just not doing what it's supposed to do. — gymnastm117
– gymnastm117, Commented Jan 21, 2015 at 19:54
@gymnastm117 He never said it didn't compile. Debugging it would have shown the unexpected value of i in the print loop. Pretty sure that was his point. — WhozCraig
– WhozCraig, Commented Jan 21, 2015 at 19:55
int a[n] should throw an error. You cannot initialize arrays in c++ with a variable size. You would need to make n a const variable. const int n = 7; int a[n]; — user2970916
– user2970916, Commented Jan 21, 2015 at 19:55

Community · Accepted Answer · 2017-05-23 12:12:49Z

1

In print and getInput your variable i is not initialized to 0

You should initialize your i to 0

for(int i = 0; i<n;i++)
{
    cout<<"Number? ";
    cin>>a[i];
}

Same for the print method.

Also, you should initialize your array size with a cont var. For more details

const int n = 7;

edited May 23, 2017 at 12:12

CommunityBot

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answered Jan 21, 2015 at 19:45

Jérôme

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2 Comments

gymnastm117 Over a year ago

Oh right thank you, I can't believe I missed that! But it still isn't asking for user input.

Tasos Vogiatzoglou Over a year ago

It's a good practice to get used to compiling with -Wall and even -pedantic -Werror. These are errors that a compiler can notify you about and you'll get a better feeling of what is standard behavior.

KidWithAComputer · Accepted Answer · 2015-01-21 20:04:42Z

0

 void print(int a[ ], int n)
 {
     for(int i; i<n; i++)
 {
    cout<<a[i]<<"    ";   
 }

    cout<<endl;
 }

This is your function, in which you have not initialize the value of i. Initialize i =0; Make it:

 for(int i = 0; i<n; i++)

answered Jan 21, 2015 at 20:04

KidWithAComputer

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