How to store an array into an array in C?

Question

Let's say I have a fixed size 3 of array that stores the RGB of color. Example:

color[3] = {0, 0, 255}

and I have another array, arrayOfColors that stores many colors. Example:

arrayOfColors = { {0, 0, 255}, {0, 0, 0}, {255, 255, 255} }

I'm not sure what is the best way to do this but I've tried something. But I have errors when I do this approach. Please help me out, I'm very new to C language. Thank you in advance!

unsigned char color1[3] = {0, 0, 0};
unsigned char color2[3] = {0, 255, 255};

unsigned char *rowColors = NULL;
rowColors = (unsigned char*)malloc((2) * sizeof(char));
rowColors[0] = color1;
rowColors[1] = color2;

Achal · Accepted Answer · 2019-08-25 11:29:33Z

1

Here

unsigned char *rowColors = NULL;

rowColors is a unsigned character pointer i.e rowColors[0] is a single character and a character can not hold character buffer like below

rowColors[0] = color1; /* rowColors[0] is a char, it can't hold char buffer color1 */
rowColors[1] = color2;

If you wish to store multiple character buffer like color1 and color2 into rowColors, it should either of unsigned char** type or array of pointers not just unsigned char* type.

Sample code

unsigned char color1[3] = {0, 0, 0};
unsigned char color2[3] = {0, 255, 255};

unsigned char *rowColors[2]; /* USE ARRAY OF POINTERS */
for(int row = 0;row < 2; row++) {
    rowColors[row] = malloc(MAX_EACH_BUF_SIZE)); /* ALLOCATE MEMORY FOR EACH POINTER FIRST, define MAX_EACH_BUF_SIZE value accordingly */
}
/* AND THEN STORE INTO ARRAY OF POINTER */
rowColors[0] = color1;
rowColors[1] = color2;

edited Aug 25, 2019 at 11:29

answered Aug 25, 2019 at 11:23

Achal

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2 Comments

Thomas Over a year ago

Thank you, this is the answer that I was looking for :), sorry I can't upvote your post because I don't have enough reputation points

Achal Over a year ago

@Thomas I'm glad about the fact that my answer helped you. You are welcome on Stack Overflow.

Kamol Hasan · Accepted Answer · 2019-08-25 12:49:04Z

1

You can do using C structure. This doesn't describe the process of declaring an array inside another array but it will serve your purpose.

You need to declare a structure type object first. Then make an array of that structure. See bellow example for better understanding:

struct RGBColor
{
  unsigned int R, G, B;
};

RGBColor arrayOfColors[] = { {0, 0, 255}, {0, 0, 0}, {255, 255, 255} };

int main ()
{
  int len = sizeof (arrayOfColors) / sizeof (arrayOfColors[0]);
  int i;
  for (i = 0; i < len; i++)
    {
      printf ("{%d, %d, %d}\n", arrayOfColors[i].R, arrayOfColors[i].G,
          arrayOfColors[i].B);
    }
  return 0;
}

Output:

{0, 0, 255}
{0, 0, 0}
{255, 255, 255}

edited Aug 25, 2019 at 12:49

answered Aug 25, 2019 at 12:34

Kamol Hasan

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3 Comments

user3629249 Over a year ago

The returned type from sizeof() is a size_t, not an int

user3629249 Over a year ago

This answer is missing the statement: #include <stdio.h>. Suggest not expecting a beginner in C to know this

Thomas Over a year ago

@kamolHasan thank you! but if I want to send the arrayOfColors with the struct RGBColor type to other processor, what datatype should I pass in? For example: MPI_Send(arrayOfColors, 3, ?DATATYPE?, 0, 0, MPI_COMM_WORLD);

John Bode · Accepted Answer · 2019-08-25 16:20:18Z

Remember that unless it is the operand of the sizeof or unary & operators, array expressions like color1 and color2 will be converted (“decay”) from expressions of type “N-element array of T” to “pointer to T”, and the value of the expression will be the address of the first element of the array. So in an expression like

x = color1;

the type of x needs to be unsigned char *. So you can start by setting up an array of unsigned char * to store the addresses of separate arrays like color1 and color2:

unsigned char *colors[2];

colors[0] = colors1;
colors[1] = colors2;

However, what might ultimately be simpler is just setting up an Nx3 array of unsigned char:

unsigned char colors[N][3];  // for some N

memcpy( colors[0], color1, sizeof color1 );

To allocate it dynamically, use

unsigned char (*colors)[3] = malloc( sizeof *colors * N );

and you can access each element like a regular array:

colors[0][0] = color1[0];

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