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Unit 19: Multi-Dimensional Arrays

At the beginning of this unit, we say that an array can hold one or more values of some type T. In fact, T can also be an array. So, we can have an array of array of long, for instance. 

long matrix[10][20];
Here, we have an array of 10 elements, and each element is an array of 20 long values.

When we access the elements in the array, we can use the notation matrix[i][j] (which is actually (matrix[i])[j]).

Such an array is called a two-dimensional array, or 2D array. We can have a 3D array, 4D array, and so on.

We have seen three types of arrays. On the stack, we have fixed length array and variable length array. On the heap, we have dynamically allocated arrays. A 2D array can mix different types of array. Since we discourage the use of variable length array, we will focus on fixed-length arrays and dynamically allocated arrays only.

Fixed Length 2D Array

The example matrix above is a fixed length array. In the memory, a continuous space of 200 long values have been allocated, and we can visualize this is having 10 rows of long array, each array contains 20 columns of long values.

With array decay, when we use the first level index of the array, matrix[i], this is equivalent to &matrix[i][0] (the address of the first element in matrix[i]), and it has the type long *. Thus, if we want to pass individulal row of a 2D array into a function, we have two options for parameters declaration:

  • void bar(long num_of_cols, long* matrix_row) { .. }
  • void bar(long num_of_cols, long matrix_row[]) { .. }

This is similar to a 1D array since matrix[i] is a 1D array.

We can then invoke the function bar like:

bar(20, matrix[i]);

Things get a bit tricky when we want to pass a 2D array into a function. By array decay, when we use the array name matrix, it is equivalent to &matrix[0], which is the address of the first element in matrix, which is actually the address of an array of 20 long.

We can declara the parameter for a function using either one of the following:

  • void qux(long num_of_rows, long num_of_cols, long (* matrix_row)[20]) { .. }
  • void qux(long num_of_rows, long num_of_cols, long matrix_row[][20]) { .. }

and call it as:

qux(10, 20, matrix);

Note the parenthesis in the declaration of long (* matrix_row)[20]. This declaration is different from long* matrix_row[20] (which is actually long *(matrix_row[20]) Let's decipher this:

  • long *(matrix_row[20]) actually means matrix_row is an array of 20 pointers to long values.
  • long (* matrix_row)[20] actually means matrix_row is a pointer to an array of 20 long values.

It is also important to note that, we cannot omit the number 20 in the [] since it is part of the type information. A pointer to an array of 20 long values is treated as a different type than a pointer to an array of 19 long values in C.

Fixed Size Array of Dynamically Allocated Array

Suppose that we know only one of the dimension of the array in advanced, but not the other dimension. We can allocate a fixed length array for the known dimension, and allocate the other dimension dynamically using calloc. We can declare an array like this:

double *buckets[10];
long num_of_cols = cs1010_read_long();
for (long i = 0; i < 10; i += 1) {
  buckets[i] = calloc(num_of_cols, sizeof(double));
}

Here, bucket is a 1D array of 10 pointers to long. So, we can easily pass bucket to a function just like any other 1D array:

  • void baz(long num_of_rows, long num_of_cols, long **bucket)
  • void baz(long num_of_rows, long num_of_cols, long *bucket[])

Access such individual elements in such type of array is no different from accessing a fixed length 2D array: we use bucket[i][j].

Remember to free the allocated memory after we are done:

for (long i = 0; i < 10; i += 1) {
  free(buckets[i]);
}

Dynamically Size 2D Array

Suppose that we do not know both dimensions in advanced, then we can allocate both dimensions of the array dynamically on the heap.

double **canvas;
long num_of_rows = cs1010_read_long();
long num_of_cols = cs1010_read_long();
canvas = calloc(num_of_rows, sizeof(double *));
for (long i = 0; i < num_of_rows; i += 1) {
  canvas[i] = calloc(num_of_cols, sizeof(double));
}

Passing such array into a function is no different from a 2D array where only one dimension is dynamically allocated above.

Remember to free the allocated memory for both dimension after we are done:

for (long i = 0; long i < num_of_rows; i += 1) {
  free(canvas[i]);
}
free(canvas);

Note that during deallocation of memory, we need to do it in the reverse order of memory allocation. If we call free(canvas) first, we are no longer guaranteed to be able to access canvas[i] with the correct pointers inside, so calling free(canvas[i]) after free(canvas) might lead to an error.

Jagged Array

One advantage of using a dynamically allocated array is that it allows a jagged 2D array, where each row has a different size. The example below allocate memory for a 2D array that is shaped like a half-square: the first row has one element, second row two elements, third row three, and so on.

double *half_square[10];
for (long i = 0; i < 10; i += 1) {
  half_square[i] = calloc(i+1, sizeof(double));
}

Initializing a Multidimensional Array

Just like a 1D-array, we can initialize a multi-dimensional array with initializers during declaration:

long matrix[3][3] = {
    {1, 0, -1},
    {-1, 1, 0},
    {0, -1, 1}
};
Note that we use nested { and } here. There are other variations to the syntax above, which you may read up on your own if you are interested as we do not need to write complex initializers for multi-dimensional arrays that often.

Problem Set

Problem 19.1

Write two functions described below. Show how you would declare the parameters to each function and how you would call each function.

a) Write a function add that performs 3x3 matrix addition. The function should operate on 3x3 matrices of long, takes in three parameters, the first two are the operands for addition and the third is the result.

b) Write a function multiply that performs 3x3 matrix multiplication. The function should operate on 3x3 matrices of long, takes in three parameters, the first two are the operands for multiplication and the third is the result.

Problem 19.2

We need to represent the distance in km between every major cities in the world. Let's label every city with a number, ranging from 0 .. n-1, where n is the number of cities. The distance between city i and j is the same as the distance between city j and i. The distance can be represented with long.

Explain how you would represent this information using jagged two-dimensional array in C efficiently. We have information of a few thousand cities to store.

Explain how you would write a function long dist(long **d, long i, long j) to retrieve the distance between any two city i and j.

Appendix: Complete Code from Lecture

include "cs1010.h"

void print_row(long ncols, long row[]) {
  for (long j = 0; j < ncols; j += 1) {
    cs1010_print_long(row[j]);
    cs1010_print_string(" ");
  }
  cs1010_println_string("");
}

void matrix_print(long nrows, long ncols, long (*matrix)[20]) {
  for (long i = 0; i < nrows; i += 1) {
    print_row(ncols, matrix[i]);
  }
}

int main()
{
  long matrix[3][20] = {
    {1, 0, 0},
    {0, 1, 0},
    {0, 0, 1}
  };

  matrix_print(3, 3, matrix);
}

#include "cs1010.h"

void print_row(long ncols, long row[]) {
  for (long j = 0; j < ncols; j += 1) {
    cs1010_print_long(row[j]);
    cs1010_print_string(" ");
  }
  cs1010_println_string("");
}

void print(long nrows, long ncols, long *matrix[10]) {
  for (long i = 0; i < nrows; i += 1) {
    print_row(ncols, matrix[i]);
  }
}

int main()
{
  long *buckets[10];
  long num_of_cols = cs1010_read_long();
  for (long i = 0; i < 10; i += 1) {
      buckets[i] = calloc(num_of_cols, sizeof(long));
  }
  print(10, num_of_cols, buckets);
  for (long i = 0; i < 10; i += 1) {
      free(buckets[i]);
  }
}

#include "cs1010.h"

void print_row(long ncols, long row[]) {
  for (long j = 0; j < ncols; j += 1) {
    cs1010_print_long(row[j]);
    cs1010_print_string(" ");
  }
  cs1010_println_string("");
}

void print(long nrows, long ncols, long *matrix[10]) {
  for (long i = 0; i < nrows; i += 1) {
    print_row(ncols, matrix[i]);
  }
}

int main()
{
  long **canvas;
  long num_of_rows = cs1010_read_long();
  long num_of_cols = cs1010_read_long();
  canvas = calloc(num_of_rows, sizeof(long *));
  for (long i = 0; i < num_of_rows; i += 1) {
      canvas[i] = calloc(i + 1, sizeof(long));
  }

  // print(num_of_rows, num_of_cols, canvas);

  for (long i = 0; i < num_of_rows; i += 1) {
      free(canvas[i]);
  }
  free(canvas);
}