# N-dimensional arrays in C

The grammar of C is messy and some constructions can be counter-intuitive for beginners: this is the case of multidimensional arrays (non-programmers would call them tensors).

## Introduction

Someone new to low-level programming would be tempted to declare arrays with multiple dimensions in the following way:

float*** make_3d_float_array(size_t x, size_t y, size_t z) {
float*** A = malloc(x*sizeof(float**));

for (...) {
A[i] = malloc(... // I think you know the rest...


Even though the syntax for accessing cells is the same as for static arrays, that's not the best thing to do: unless you're planning to replace entire rows or planes more often than you do random accesses, this is inefficient. We need contiguous arrays without having to explicitly compute the indexes.

While learning C and experimenting with metaprogramming, I wrote a few helpers using the preprocessor for dynamic multidimensional arrays. If you take a look at it without much knowledge of the C preprocessor, you'll probably see it as mostly voodoo. Fortunately C99 comes with syntactic sugar that makes this unnecessary. Read on !

## Static arrays

As a reminder, static arrays means we are putting the size of the array in its type. A static 2D arrays of integers can be declared and used as such:

void foo(int x[4][2]) {
// Do something with x.
}

int main() {
int x[4][2];

for (int i = 0; i < sizeof(x) / sizeof(x[0]); i++)
for (int j = 0; j < sizeof(x) / sizeof(x[0][0]); j++)
x[i][j] = rand();

foo(x);
// Don't be fooled by the syntax of foo's argument,
// there's no copy of x involved!

return 0;
}


Although the syntax may be deceiving, the array is not copied when calling the function. And there's no way to ask for an implicit copy.

What about dynamic arrays, for which the size may not be known at compile time ?

## Proper dynamic arrays

In C99, variable length arrays can be declared and allocated as follow:

int m = rand() % 10 + 1;
int n = rand() % 10 + 1;

printf("m = %d, n = %d\n", m, n);

int x[m][n];
int (*y)[n] = malloc(m * n * sizeof(y[0][0]));
int (*z)[m][n] = malloc(sizeof(*z));


x is a dynamic (i.e. variable-length) array allocated on the stack. y and z are both pointers to chunks of memory allocated in the heap (which are the size of x) but the semantic slightly differs between the two: the type of the data y points to is an array of n integers, while z points to an array of m arrays of n integers.

If you're not comfortable with pointer arithmetic, you may want to read this tutorial.

Cell accesses are written as such:

x[i][j] = 42;
y[i][j] = 42;
(*z)[i][j] = 42; // same as z[0][i][j] ;)


Even though the semantic of z is, in some way, closer to what we're expressing, its usage is less practical.

How can we write functions taking dynamic arrays as arguments ? This is were the real syntax tricks hide: we can put the lengths as arguments and use them in the type of the array. For more details on it, read this GCC man page.

In plain C99:

void foo(int m, int n, int x[m][n]) {
// Do something with x.
}

// ...

foo(m, n, x);
foo(m, n, y);
foo(m, n, *z);


Again, the arrays are not copied.

Finally, GCC provides an elegant extension to put the size arguments after the array, which may lead to more readable code in some cases:

void bar(int m, int n; int x[m][n], int m, int n) {
// Do something with x.
}

// ...

bar(x, m, n);


A full working example of this is available here.