如何在C中在另一个结构内使用结构（同时使用malloc和2D数组？）

Question

I am new to using struct, hope you can help me. I want to have a nested struct such that a member of A is type B struct. The code I have so far is shown below.

typedef struct B
{
   int b1;
} b;

typedef struct A
{
    b a1;
}a;

But I also need 'a1' to be a dynamic 2d array, so I did the following code (not sure if this is right as i used a pointer to a pointer for this in order for the 'array' size to be set at runtime):

typedef struct B
{
   int b1;
} b;

typedef struct A
{
    b **a1;
}a;    

To store memory I used malloc(didnt include the error detection part). I need to refer and compare each member of a1 by x, shown below (where a and b are integers). Can I do this?:

a* one;
b** two;

one= malloc(10* sizeof(a*));
two= malloc(10* sizeof(b*));
for(x = 1; x <= 10; x++)
    {
        two[x] = malloc(10* sizeof(b));
    }

 if(one[x].two[a][b] == x)
    one[x].two[a][b].b1 =2;

Sorry if my code doesn't make sense at all.

Answer 1

I also need 'a1' to be a 2d array, so I did the following code (not sure if this is right):

 typedef struct B { int b1; } b; typedef struct A { b **a1; }a; 

The code provides valid C declarations, but there is no 2d array in sight. Member a1 of struct A is a pointer to pointer to struct B , which is not at all the same thing. If you truly want to declare a 2d array then you must give its bounds, for instance

typedef struct A {
    b a1[5][7];
} a;

If you don't know the needed sizes until runtime, however, then the double pointer is indeed what you want -- just do not confuse it with a 2D array.

To store memory I used malloc(didnt include the error detection part). I need to refer and compare each member of a1 by x, shown below (where a and b are integers).

Well no, a and b cannot be integers because you're already using them as type names in the same scope.

Can I do this?:

 a* one; b** two; one= malloc(10* sizeof(a*)); two= malloc(10* sizeof(b*)); for(x = 1; x <= 10; x++) { two[x] = malloc(10* sizeof(b)); } 

In addition to the name collision already mentioned, your allocation of one is wrong. It should be ...

 one = malloc(10 * sizeof(a));

... or even better ...

 one = malloc(10 * sizeof(*one));

. Your allocations for two and two[x] are ok, but they would be better if rewritten in the same manner as my latter example above.

  if(one[x].two[a][b] == x) one[x].two[a][b].b1 =2; 

The type of one[x].two[i][j] (for integer i and j ) is b , which is to say struct B . The type of x is int . These are not comparable, so no, you cannot evaluate the expression in the if condition. The assignment to one[x].two[a][b].b1 is ok, though.

Answer 2

Instead of messing around with 2D dynamic memory allocation, you can fake a 2D array with a 1D array.

b *a1;
a1 = malloc(sizeof(b) * width * height);
for(i = 0; i < height; i++)
    for(j = 0; j < width; j++)
        do_stuff(a1[i*width + j];  //simulates a1[i][j]

The only thing this requires is that you store your width and height values somewhere, eg

typedef struct A {
    b *a1;
    int height;
    int width;
}