How to define a double dimensional array using a single pointer

Question

is char a[64][] equivalent to char *a[64]?

If yes, what if I want to declare char a[][64] using a single pointer. How can i do it?

Answer 1

char a[][64]

is similar in pointer representation to

char (*a)[64]

you can read this. for 2D-array the second dimension size should be specified.

Answer 2

You're looking for a pointer to an array:

char (*a)[64];

Answer 3

Perhaps you are not sure about how the following two statements are different.

char* a[64];
char (*a)[64];

The first one defines a to be an array of 64 char* objects. Each of those pointers could point to any number of chars. a[0] could point to an array of 10 chars while a[1] could point to an array of 20 chars. You would do that with:

a[0] = malloc(10);
a[1] = malloc(20);

The second one defines a to be a pointer to 64 chars. You can allocate memory for a with:

a = malloc(64);

You can also allocate memory for a with:

a = malloc(64*10);

In the first case, you can only use a[0][0] ...a[0][63]. In the second case, you can use a[0][0] ... a[9][63]

Answer 4

char a[64][] is equivalent to char *a[64].

No, because char a[64][] is an error. It attempts to define a to an array of 64 elements where each element is of type char[] - an incomplete type. You cannot define an array of elements of incomplete type. The size of element must be a fixed known constant. The C99 standard §6.7.5.2 ¶2 says

The element type shall not be an incomplete or function type.

Now if you were to compare char a[][64] and char *a[64], then again they are different. That's because the array subscript operator has higher precedence than *.

// declares an array type a where element type is char[64] - 
// an array of 64 characters. The array a is incomplete type
// because its size is not specified. Also the array a must have
// external linkage.
extern char a[][64];

// you cannot define an array of incomplete type
// therefore the following results in error.
char a[][64];

// however you can leave the array size blank if you 
// initialize it with an array initializer list. The size
// of the array is inferred from the initializer list.
// size of the array is determined to be 3
char a[][2] = {{'a', 'b'}, {'c', 'd'}, {'x', 'y'}};

// defines an array of 64 elements where each element is 
// of type char *, i.e., a pointer to a character
char *a[64];

If you want to declare a pointer to an array in a function parameter, then you can do the following -

void func(char a[][64], int len);
// equivalent to
void func(char (*a)[64], int len);

char (*a)[64] means a is a pointer to an object of type char[64], i.e., an array of 64 characters. When an array is passed to a function, it is implicitly converted to a pointer to its first element. Therefore, the corresponding function parameter must have the type - pointer to array's element type.