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carraysstruct
Garde Des Ombres
Garde Des Ombres

Reputation: 191

Defining a 2D array of structs type in C

I have the following struct:

typedef struct Position Position;
struct Position
{
    int x; 
    int y; 
};

I want to define a new type which is a 2D array of the Position struct . How can i do that? and what's the type of the function that returns such array ?

Upvotes: 0

Views: 140

Answers (2)

RobertS supports Monica Cellio
RobertS supports Monica Cellio

Reputation: 15062

I want to define a new type which is a 2D array of the Position struct. How can I do that?

You can typedef a 2D-array of struct Position:

typedef struct Position
{
    int x; 
    int y; 
} Position;
typedef Position Postion2d[M][N];

where M is the amount of elements in the first dimension and N is the amount of elements in the second dimension.

Online Example:

typedef struct Position
{
    int x; 
    int y; 
} Position;

typedef Position Postion2d[4][5];

int main()
{
    Postion2d a; 
    a[0][1].x = 2;
    a[0][1].y = 5;
}

Upvotes: 2

Hitokiri
Hitokiri

Reputation: 3699

You can declare the array of struct Position as the array of the other primitive types (for instance, int, float, etc). You can also use the double pointer.

Position **p;
Position p[100][100]

My example below using the double pointer to manipulate the size of list of position.

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

typedef struct 
{
    int x; 
    int y; 
}Position;

Position ** return_array(int m, int n) {
    Position **p = malloc (sizeof(Position *) * m);
    for (int i = 0; i < m; i++) {
        p[i] = malloc(sizeof(Position)*n);
        for (int j = 0; j < n; j++) {
            p[i][j].x = i;
            p[i][j].y = j;
        }
    }
    return p;
}

int main()
{
    Position **pp = return_array(4, 4);
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            printf("P[%d][%d].(x,y) = (%d, %d)\n", i, j, pp[i][j].x, pp[i][j].y);
        }
    }
    return 0;
}

The result:

P[0][0].(x,y) = (0, 0)
P[0][1].(x,y) = (0, 1)
P[0][2].(x,y) = (0, 2)
P[0][3].(x,y) = (0, 3)
P[1][0].(x,y) = (1, 0)
P[1][1].(x,y) = (1, 1)
P[1][2].(x,y) = (1, 2)
P[1][3].(x,y) = (1, 3)
P[2][0].(x,y) = (2, 0)
P[2][1].(x,y) = (2, 1)
P[2][2].(x,y) = (2, 2)
P[2][3].(x,y) = (2, 3)
P[3][0].(x,y) = (3, 0)
P[3][1].(x,y) = (3, 1)
P[3][2].(x,y) = (3, 2)
P[3][3].(x,y) = (3, 3)

Upvotes: 5

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