pointer and multidimensional array

Question

i want to print the value which contain in the multidimensional array.I know that if we write something like arr[3][4] then arr[0] is the pointer to the first element in the that multidimensional array. I want to ask why this code give me garbage value after the program print the number containing in the multidimensional array?

this is the code:

    #include <stdio.h>

    #define ROW 3
    #define COLL 4

    int main(void) {
    int arr[ROW][COLL]={{1,2,3,4},
                  {5,6,7,8},
                  {9,10,11,12}};
    int *ptr;

    for(ptr=arr[0];ptr<arr[ROW]+COLL;ptr++){
        printf("%d ",*ptr);
    }
    printf("\n");
    return 0;
    }

this is the result when i compile the above code:

1 2 3 4 5 6 7 8 9 10 11 12 -1079481492 134513936 0 -1079481352

but after changing the for loop to the following:

for(ptr=arr[0];ptr<=arr[ROW-1]+COLL-1;ptr++)

the code work and give the exact number which contain in the multidimensional array.

array access row from 0..1..and 2 and here you want to access 3 means fourth raw's first element which is not there. So it invokes undefined behavior. — Jayesh Bhoi
– Jayesh Bhoi, Commented Apr 23, 2014 at 13:47

Filipe Gonçalves · Accepted Answer · 2014-04-23 13:36:52Z

2

Because arr[ROW] is an out-of-bounds access. The last valid position in arr is arr[ROW-1]. Thus the first version of your code invokes undefined behaviour.

answered Apr 23, 2014 at 13:36

Filipe Gonçalves

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2 Comments

harianja Over a year ago

what is out-of-bound acces? can you explain that?

John Bode Over a year ago

@harianja: He means that you're trying to access something past the end of the array.

SzG · Accepted Answer · 2014-04-23 13:38:58Z

0

The pointer to the last element is

arr[ROW-1] + COLL - 1

So you loop should be

    for (ptr = arr; ptr < arr[ROW-1] + COLL; ptr++) {

answered Apr 23, 2014 at 13:38

SzG

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Comments

user3478972 · Accepted Answer · 2014-04-23 14:27:19Z

Elements of an array are indexed from 0 so the last element in your matrix is arr[ROW-1][COLL-1].

When your pointer reaches arr[2] + COLL there is no such element because the second coordinate of the last element in last row is [COLL-1].

Elements of an array are stored in 1D so arr[0] + 4 is acctualy arr[1][0]. When you increment ptr++ it tells the compiler to access the next memory location after the current. So, when ptr points to a[0] + 4 it works because the compiler access the fifth value after a[0][0] which is actually the first value in the next row, arr[1][0], because their memory locations are next to each other.

When your ptr points to arr[4] + i, i = 0, 1, 2, 3 you get bad output because those elements are not in your array.

nishantbhardwaj2002 · Accepted Answer · 2014-04-23 14:45:10Z

Because your first element, 1, is at arr[0]+0,

5 is at arr[1]+0,

9 is at arr[2]+0,

10 is at arr[2]+1,

11 is at arr[2]+2,

and 12, which is the last element of your array is at arr[2]+3.

but your for loop keeps moving even after the array ends, that is, until the memory location arr[3]+4 is reached. Giving you four garbage values because we have stored nothing in the memory after arr[2]+3.

so the for loop should be like:

for (ptr=arr; ptr <= arr[ROW-1]+COLL-1; ptr++);

which is equivalent to:

for (ptr=arr[0]; ptr <= arr[ROW-1]+COLL-1; ptr++);

as arr[0] and arr point to the same location in memory. NOTE- try reading how memory is allocated in arrays of multiple dimensions.

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