How can you determine the address of a two dimensional array element in c?

Answer 1

If the array is a local fixed-length array, use the address-of operator (&).

Example:

int x[3][4]; // local fixed-length array

int* p = &x[2][1]; // obtain the address of element x[2][1]

//...

For variable-length arrays we need to use a pointer and separate variables to keep track of the dimensions. C uses row-major semantics so the first dimension specifies the number of rows. However, the type of element in each row is not int, it is int[4], as per the second dimension. In other words, it is one-dimensional array of 3 one-dimensional arrays each of length 4. However, we cannot have a pointer of type int[4] because int[4] also decays to a pointer. We could use a pointer to pointer to int (int**), however in order to allocate the array contiguously without having to maintain a separate array of pointers into the second dimension, it is simpler to treat the array as if it were one-dimensional using a pointer to int (int*). After all, the actual elements we wish to access are integers, so that's what we should really be referring to, rather than to rows of arrays of integers which adds an unnecessary level of indirection. This then means we must use pointer-arithmetic to determine the address of each element. The following demonstrates how to use pointer arithmetic to access each element of a two-dimensional array through a pointer given only the row and column of the element we wish to access:

const size_t rows = 3;

const size_t cols = 4;

int* array2d = malloc (rows * columns * sizeof (int)); // allocate memory to the array

for (int row=0; row<rows; ++row) {

for (int col=0; col<cols; ++col) {

int* p = array2d + (row * cols) + col;

// ...

}

}

free (array2d); // release allocated memory

Here, row * cols determines the offset address of the start of the given row. array2d refers to the start address, so adding array2d gives us the actual address of the row. Adding col then gives us the address of the element within the given row. Note that array2d, &array2d and &array2d[0] are equivalent; they all refer to the start address of the array.

The same technique can also be used when passing a two-dimensional array to a function:

void f (int* array2d, const size_t rows, const size_t cols) {

for (size_t row=0; row<rows; ++row) {

for (size_t col=0; col<cols; ++col) {

int* p = array2d + (row * cols) + col;

// ...

}

}

}

Answer 2

2D arrays in C and C++ are like Excel spreadsheets. The data contained in a specific cell is found using its row and column.

For example:

int table[3][3];

for (int i = 0; i < 3; i++)

{

for (int j = 0; j < 3; j++)

{

table[i][j] = (rand()+time(0))%10;

}

}

This code will create a 3x3 2D array named "table" with random 0-9 numbers in each spot.

Back to the question using this example:

To pull information from one specific position, you would use:

table[i][j]

where 'i' and 'j' can be switched out for a specific number or variable

The question was posted in C++, so you may need to modify the code for any non-C compatible stuff from my C++ code.