How is 2D array passed to a function having formal parameter as a pointer to a array of elements ?

Question

when I am passing a 2D array and function is declared like
func(int a[][6]) //This implies that a is going to an array of 6 columns ,rght now we don't know the rows fine ,but it is pointing to a 2D array .

but if I declare the function like

func(int (*a)[6]) //Now how is this notation correct for a to point to a 2D array since it implies that a is a pointer to an array of 6 elements so that implies that it would be pointing to a 1D array .Although we can increment it and get to another row of 2D array ,but still I am a bit confused with this notation .

I am unable to get the exact way in this pointer a would be pointing to the 2D array .

 

 

 

 

Comments

Whenever an array is being passed as a, it decays to the address of its first element. In the case of a 1D array, if we pass a, it gets internally decayed to &a[0]

When a function has an argument like this int a[][6], internally it decays &a[0] // which in this case is the address of the first row of the 2D array, now for the function to catch this, it needs a pointer that points to an array of 6 elements (one row) i.e. int (*a)[6].

So internally it is always int (*a)[6], but for convenience and better readibility, we are writing as int a[][6].

In both case, a is a pointer to an array, which points to the first row, now a+1 points to the next row, *(a+1) derefences to address of the first element, *(*(a+1)) represents the value of the first element

Now to pass the whole 2D array, we can have an argument like int (*a)[5][6] and pass it like &a which represents the whole 2D array and does not decay to the first row.

Answer 1

Hi 

we can write a[i] as *(a+i) .

as [] is left to right Associativity . expand it from left to right

a[i][j] can be written as ( *(a+i)) [j] = *( *(a+i) +j)  

when we pass an array , we pass address of first element in array, i.e a[0][0]

a[0][j] can be written as ( *(a+0))[j] = (*a)[j]

if we increment a then it will become (*(a+1))[j]

Answer 2

func(int a[][6])  // This implies that a is going to an array of 6 columns ,right now we don't know the rows fine ,but it is pointing to a 2D array .

I hope you are fine with this declaration. Here a will be having the address of first element a[0][0].

func(int (*a)[6]) //Now how is this notation correct for a to point to a 2D array since it implies that a is a pointer to an array of 6 elements so that implies that it would be pointing to a 1D array .I am unable to get the exact way in this pointer a would be pointing to the 2D array .

 

As you know passing an array is just sending Base address.

Now in given declaration "int (*a)[6]", as you have mentioned you can access the each row of 6 integers base address.Once you will do (a+1) it will increment to next row base address and so on.

To access the each element of that row you have to just do like below

 

int *t;

t = a;   // As you know a is a pointer to an array of 6 elements.

           // It means it can have address to such array of 6 elements.

          //  we are assigning that base address to a integer pointer.

          // Since integer pointer can hold the address of integer element.

          // (i.e. first element of first array)

int i;

for( i =0;i<6;i++)

   printf("%d", *(t+i);

 

This way you can access each elements of the 2-D array.

 

Comments

But this way I can access one row , how to read all the rows ,there' where am I stuck .

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Whenever you store a array base address into a pointer like below

int b[] = {1,2,3,4,5};

int *c = b;

and now when you will call (c+1), it is actually the next object of type c.

Lets take b = 100 and i am assuming integer will take 2 Byte in memory and also the memory is byte addressable.So c will have 100 after assigning b to it.So once you will call (c+1) it will be actually 102.which is nothing but the next object of type c (i.e. integer).

In your second scenario, you have a is pointer to array of 6 integer. Here also once you will call (a+1), it will give the address to the next object of type a (i.e. next pointer to array of 6 integers).

Lets take a = 1000 and here also i am assuming integer will take 2 Byte in memory and also the memory is byte addressable. So this case array of 6 integer will take 12 Byte.So once you will call (a+1) it will be actually 1012.which is nothing but the address of first element of second row.

Once you got the address of first element of array , you can traverse the all element of that array in the same way.