Calling a sort function on an array c++

Question

I am using an insertion sort function to sort a 5000 int ascending array. when i pass the array's address to the function call, i get [Error] name lookup of 'i' changed for ISO 'for' scoping [-fpermissive]. Maybe I should use a vector but I am unfamiliar with them. Maybe I coded something wrong?

#include <iostream>
#define SIZE 5000 //array size
using namespace std;

void insertionSort(int arr[], int length) {
      int i, j, tmp;
      for (i = 1; i < length; i++) {
            j = i;
            while (j > 0 && arr[j - 1] > arr[j]) {
                  tmp = arr[j];
                  arr[j] = arr[j - 1];
                  arr[j - 1] = tmp;
                  j--;
            }
      }
}

int main() {
    int a[SIZE];
    int count = 1;

    for(int i=0; i<SIZE; i++) {
        a[i] = count;
        count++;
    }

    for(int i = 0; i < SIZE; i++) {
        printf("%d\t", a[i]);
    }

    insertionSort(&a[i], SIZE);
}

In your main() function i is not defined when you call insertionSort(&a[i], SIZE);. This is because i only exists inside of your for loops... See @michael-schmitt answer below. — nonsensickle
– nonsensickle, Commented May 5, 2015 at 0:15

Michael Schmitt · Accepted Answer · 2015-05-05 00:04:50Z

2

You probably want to call

 insertionSort(a, SIZE)

answered May 5, 2015 at 0:04

Michael Schmitt

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2 Comments

Robert Over a year ago

Wow. Thank you. I thought you could only pass array arguments by reference. I dont just want a fix, could you explain why i didn't have to pass the address instead of just the array? And, thank you again for the advice.

acenturyandabit Over a year ago

The array identifier points to the start of the array, so it is essentially the address already.

Andreas DM · Accepted Answer · 2015-05-05 09:58:34Z

2

// insertionSort(&a[i], SIZE); Here you haven't declared i in funciton main(), only in your for loop.
And the call to insertionSort is outside of your loop.

Passing (&a)[0] and a mean the same thing.

edited May 5, 2015 at 9:58

answered May 5, 2015 at 0:12

Andreas DM

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2 Comments

Robert Over a year ago

Thank you all for your time.

CiaPan Over a year ago

'Passing (&a)[0] and a mean the same thing.' – that's wrong. Should be: 'Passing &(a[0]) and a mean the same thing.'

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