What is a C-Array?

A reference to the beginning of consecutive memory of a given type

int numbers[] = {32, 45, 12, 39, 7, 15};

Purpose of C-Arrays

• Arrays allow storing many elements of the same type
• These elements can be passed to a function together
• The number of elements does not have to be known upfront (only C99)

Which Problems do C-Arrays Cause?

• Arrays only store the memory address of their beginning
• Passing arrays to functions requires passing their size as well
• The memory address an array variable references cannot be changed
• Not all C compilers support dynamically sized arrays on the stack
• Arrays cannot be directly returned from functions

Creating an Array of Constant Size

#include <stdio.h>

int main() {
  unsigned int numbers[3];  // number of elements known at compile time
  numbers[0] = 15;
  numbers[2] = 5;
  numbers[1] = 10;
  printf("{ %d, %d, %d }", numbers[0], numbers[1], numbers[2]);
  return 0;
}

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Create Dynamically Sized Array

... supported by compilers that fully implement the C99 standard.

Other compilers require the use of dynamic memory management.

#include <stdio.h>

int main() {
  size_t count = 5;  // entered by the user
  unsigned int squares[count];
  for (size_t i = 0; i < count; ++i) {
    squares[i] = i * i;
  }
  for (size_t i = 0; i < count; ++i) printf("%u\n", squares[i]);
  return 0;
}

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Determine the Number of Elements

The compiler can determine the number of elements if the array was declared in the current scope.

The sizeof operator returns the total memory in bytes.

#include <stdio.h>

int main() {
  int values[] = {-5, 39, 13, -21};  // compiler determines number of elements
  printf("used memory: %ld\n", sizeof(values));
  printf("number of elements: %ld\n", sizeof(values) / sizeof(int));
  return 0;
}

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Pass the Number of Elements When Passing an Array to a Function

#include <stdio.h>

void print_info(int values[], size_t count) {  // values is actually a pointer
  // sizeof returns the size of the pointer (64 bits on modern systems)
  printf("sizeof(values): %ld\n", sizeof(values));
  printf("sizeof(values) / sizeof(int): %ld\n", sizeof(values) / sizeof(int));
  // only way to know the number of elements is a separate argument
  printf("number of elements: %ld\n", count);
}

int main() {
  int values[] = {-5, 39, 13, -21};  // compiler determines number of elements
  printf("sizeof(values): %ld\n", sizeof(values));
  printf("sizeof(values) / sizeof(int): %ld\n", sizeof(values) / sizeof(int));
  print_info(values, sizeof(values) / sizeof(int));
  return 0;
}

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Arrays Are Not Assignable

#include <stdio.h>

int main() {
  int first[] = {1, 2, 3};
  int second[] = {4, 5, 6};
  first = second;  // compiler error: array type is not assignable
  first = {7, 8, 9};  // also a compiler error
  printf("{ %d, %d, %d }", first[0], first[1], first[2]);
  return 0;
}

It is not possible to directly return an array from a function.

Its memory would be on the stack and the returned reference invalid.

Pass Array as an Argument

#include <time.h>
#include <stdlib.h>
#include <stdio.h>

void read_values(int values[], size_t count) {
  srand(time(NULL));  // Initialize random number generator
  for (size_t i = 0; i < count; ++i) {
    values[i] = rand();  // assume the values are read from user input
  }
}

int main() {
  int values[3];  // number of elements known at compile time
  read_values(values, 3);
  printf("{ %d, %d, %d }\n", values[0], values[1], values[2]);
  return 0;
}

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