I was wondering something about the multidimensional arrays and then I got a doubt about a concept, specifically about their representation on memory.

For example, an array can be defined as follows.

int a[3];

And this is represented as this.

enter image description here

Where each element is an int and all of them are in a continuous memory space.

Then a multidimensional array can be define as this.

int a[2][4];

Which is typically represented as.

enter image description here

But a more realistic representation of this array in memory wouldn't be something like this? (by analogy with the first image).

enter image description here

Because I was thinking that if, for example, int a[5]; defines 5 integers in a space of continuous memory, then int a[5][10]; defines 10 arrays of ints in a space of continuous memory, where each array define 5 integers, then any N-dimensional static array in C should be represented only as a strip of memory.


3 Answers 3


You're nearly right, yes. In fact,

int a[5][10];

declares 5 arrays each of which contain 10 integers, rater than the other way round. Other than that you're completely right.

See http://en.m.wikipedia.org/wiki/Row-major_order for a more detailed description and an explanation of why the order is important.

  • Downvote: You've provided the start of a good answer but a link to a website is not an answer. Dec 14, 2014 at 11:46
  • @jamessnell the material linked goes beyond the original question asked, so is not really part of the answer.
    – Jules
    Dec 14, 2014 at 12:39

In addition to Jules' answer, I would like to add a bit of historical perspective.

The counter-intuitiveness of this in C is due to the simple fact:

  • In general, int a[] (length unspecified) is some sort of syntactic sugar in place of int* a
  • Likewise, int a[][] (again, lengths unspecified) is some sort of syntactic sugar in place of int** a.

Because the two-level pointer must be dereferenced in that order, it determines how a int a[M][N] would have to be laid out in memory.


Here's some code that clearly shows memory layout for arrays. Using a union, for which memory is overlaid for each union member, we can see that values are stored in row-major order.

#include <stdio.h>

#define N       4*4*4
#define N_2D    8
#define N_3D    4

union multi_d_arrays {
    int one_d[N];
    int two_d[N_2D][N_2D];
    int three_d[N_3D][N_3D][N_3D];

int main()
    int i, j, k;
    union multi_d_arrays m;

    /* initialise via one_d with sequential ints */
    for (i=0; i<N; i++)
        m.one_d[i] = i+1;

    /* in all cases the output should be sequential */
    for (i=0; i<N; i++)
        printf("m.one_d[%d] = %d\n", i, m.one_d[i]);

    for (i=0; i<N_2D; i++)
        for (j=0; j<N_2D; j++)
            printf("m.two_d[%d][%d] = %d\n", i, j, m.two_d[i][j]);

    for (i=0; i<N_3D; i++)
        for (j=0; j<N_3D; j++)
            for (k=0; k<N_3D; k++)
                printf("m.three_d[%d][%d][%d] = %d\n", i, j, k, m.three_d[i][j][k]);

    return 0;
  • this doesn't even attempt to answer the question asked, "a more realistic representation of this array in memory wouldn't be..." etc. Asker appears to be well aware about how values are really laid
    – gnat
    Dec 14, 2014 at 12:20
  • @gnat - well in that case I have completely misunderstood the point of the question. Could have sworn that the OP shows memory laid out in column-major order, in which case I have to disagree with the OP's idea that that is a more realistic representation of memory. The intent of the code in this answer is to show that.
    – mhawke
    Dec 14, 2014 at 13:08

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