Array increment positioning with respect to indexer in C - array[i]++ vs array[i++]

Question

What is the difference between array[i]++ (increment outside brackets) and array[i++] (increment inside brackets), where the array is an int array[10]?

score 31 · Accepted Answer · edited Feb 20 '20 at 16:56

31

int a[] = {1, 2, 3, 4, 5};
int i = 1; // Second index number of the array a[]
a[i]++;
printf("%d %d\n", i, a[i]);
a[i++];
printf("%d %d\n", i, a[i]);

Output

1 3
2 3

a[i]++ increments the element at index i, it doesn't increment i. And a[i++] increments i, not the element at index i.

edited Feb 20 '20 at 16:56

IHS

3
1
5

answered Sep 29 '11 at 09:38

taskinoor

45,586
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2

What's missing from this answer is what `a[i]++` and `a[i++]` evaluate to. – David Heffernan Sep 29 '11 at 09:56
It is a little bit confusing because the array *values* are in about the same range as the array *indexes*. Can you offset the values (say, by 60), so there isn't any overlap between indexes and values? – Peter Mortensen Jan 13 '20 at 03:38
2

The example output is misleading, since the access of `a[i++]` in the second statement actually returns the value at index `1`, which has been inceremented to `3` before. Be aware that `x = a[i++];` is equivalent to the statement sequence `x = a[i]; i++;` So i is used as index *before* it is incremented. – Richard Neumann Jan 24 '21 at 18:17

David Heffernan · Answer 2 · 2011-09-29T09:45:01.963

array[i]++ increments the value of array[i]. The expression evaluates to array[i] before it has been incremented.
array[i++] increments the value of i. The expression evaluates to array[i], before i has been incremented.

An illustration.

Suppose that array contains three integers, 0, 1, 2, and that i is equal to 1.

array[i]++ changes array[1] to 2, evaluates to 1 and leaves i equal to 1.
array[i++] does not modify array, evaluates to 1 and changes i to 2.

A suffix operators, which you are using here, evaluates to the value of the expression before it is incremented.

score 5 · Answer 3 · answered Sep 29 '11 at 09:36

5

array[i]++ means ( *(array+i) )++. --> Increments the Value.

array[i++] means *( array + (i++) ). --> Increments the Index.

answered Sep 29 '11 at 09:36

Sadique

22,572
7
65
91

score 2 · Answer 4 · edited Feb 20 '20 at 20:22

2

Here the Array[i]++ increments the value of the element array[i], but array[i++] increments the i value which effects or changes the indication of the array element (i.e. it indicates the next element of an array after array[i]).

edited Feb 20 '20 at 20:22

Cardinal System

2,749
3
21
42

answered Sep 29 '11 at 10:03

Gouse Shaik

340
1
2
15

score 1 · Answer 5 · answered Apr 25 '20 at 18:18

1

Let's say we have this example, array[i++] = x[m++]. This means that first set array[i] = x[m] then increase the indexes like i + 1, m + 1.

answered Apr 25 '20 at 18:18

mel

19
3

score 0 · Answer 6 · edited Feb 20 '20 at 18:05

Here array[i++] increments the index number.
On the contrary, array[i]++ increments the data value of i index.

Code Snippet:

#include <iostream>
using namespace std;

int main() 
{
    int array[] = {5, 2, 9, 7, 15};

    int i = 0;

    array[i]++;
    printf("%d %d\n", i, array[i]);

    array[i]++;
    printf("%d %d\n", i, array[i]);

    array[i++];
    printf("%d %d\n", i, array[i]);

    array[i++];
    printf("%d %d\n", i, array[i]);

    return 0;
}

score 0 · Answer 7 · edited Apr 26 '23 at 12:14

#include<iostream>
using namespace std;
int main()
{
    int arr[]={1,2,37,40,5,7};
    int i = 3;
    arr[i]++;
    cout<<i<<" "<<arr[i]<<endl;
    arr[i++];
    cout<<i<<" "<<arr[i]<<endl;
    return 0;
}

Output:

3 41
4 5

In this example i = 3, so arr[3] = 40. It then increases the value from 40 to 41 .So arr[i]++ increments the value of this particular index and a[i++] first increments the index and then gives the value for this index.

Array increment positioning with respect to indexer in C - array[i]++ vs array[i++]

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