How to use pointer expressions to access elements of a two-dimensional array in C?

Question

I know that for single-dimensional arrays x=a[i] is equivalent to x=*(a+i), but how can I access elements of a two-dimensional arrays using pointers?

Answer 1

Summary: If you have a multidimensional array defined as int [][], then x = y[a][b] is equivalent to x = *((int *)y + a * NUMBER_OF_COLUMNS + b);

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Boring Details:

The (int *) cast of y above deserves some explanation, as its necessity may not be at-first intuitive. To understand why it must be there consider the following:

1. Typed pointer arithmetic in C/C++ always adjusts the typed pointer value (which is an address) by the size of the type in bytes when adding/subtracting/incrementing/decrementing by scalar.

2. The fundamental type of a multi-dimensional array declaration (not the element type; the variable type) is an array-type of one-less dimension than the final dimension.

The latter (#2) of these really needs an example to solidify. In the following, variables ar1 and ar2 are equivalent declarations.

int ar1[5][5]; // an array of 5 rows of 5 ints.  typedef int Int5Array[5];  // type is an array of 5 ints Int5Array ar2[5];          // an array of 5 Int5Arrays. 

Now the pointer arithmetic part. Just as a typed structure pointer can be advanced by the size of the structure in bytes, so can a full dimension of an array be hopped over. This is easier to understand if you think of the multi-dimensioned array as I declared ar2 above:

int (*arptr)[5] = ar1; // first row, address of ar1[0][0]. ++arptr;               // second row, address of ar[1][0]. 

All of this goes away with a bare pointer:

int *ptr = ar1; // first row, address of ar1[0][0]. ++ptr;          // first row, address of ar1[0][1]. 

Therefore, when doing the pointer arithmetic for two-dimensional array, the following would NOT work in getting the element at [2][2] of a multi-dimensioned array:

#define NUMBER_OF_COLUMNS   5 int y[5][NUMBER_OF_COLUMNS]; int x = *(y + 2 * NUMBER_OF_COLUMNS + 2); // WRONG 

The reason is hopefully obvious when you remember that y is an array of arrays (declaratively speaking). The pointer arithmetic of adding the scaler (2*5 + 2) to y will add 12 rows, thereby computing and address equivalent to &(y[12]), which is clearly not right, and in fact, will either throw a fat warning at compile time or outright fail to compile altogether. This is avoided with the cast of (int*)y and the resulting type of the expression being based on an bare pointer-to-int:

#define NUMBER_OF_COLUMNS   5 int y[5][NUMBER_OF_COLUMNS]; int x = *((int *)y + 2 * NUMBER_OF_COLUMNS + 2); // Right!