Setup pointer to 2D array, then passing that pointer into a function

Question

G'day all,

I know this has been discussed a lot, but I still cant find what I need. I am pretty new to C and still am getting my head around pointers, namely, pointers to multi-dimensional arrays.

I have looked at examples such as int a[2][3]; int (*p) = a when a is a 2D array, but what does the brackets do?

I am creating a program which has a 2D array as a variable and it needs to pass that array onto an external function to modify it.

My array is initilised like this:

unsigned int node[3][2]={ {PINB,PINC}, {0,0}, {0,0} };

I believe the best way to do this is through pointers but I don't know how to setup the prototype for my function that takes in the 2D array.

This is the function im passing it to and how I call it:

NodeToMIDI(node, i, pMIDIdata);

And this is its prototype: //where "node_pointer" is the 2D array arg.

void NodeToMIDI(unsigned int node_pointer, unsigned int node_select, unsigned int * MidiPacket);

Can someone please explain the syntax and logic behind how to do this. Some of the other threads I have looked at have syntax that is unknown to me. Such as int ** a for example.

Thank-you for any help! Andrew.

asaelr · Accepted Answer · 2012-02-23 13:49:25Z

2

You can pass the array to function naturally only if you know (at compilation-time) the size of the inner array.

The examples you saw, are probably int a[2][3]; int (*p)[3] = a. The brackets means that p is pointer to array of 3 ints, not array of 3 pointers to int. Since you can dereference pointers like arrays, you can use p as if it was an array of arrays of 3 ints (which a is).

In the same way, you can write a prototype like void myfunc(int (*arr)[3]); - If you pass a to this function, it's exactly the same thing like pass array to function that gets pointer.

There is another way to write this prototype: void myfunc(int arr[][3]); - The compiler will understand it in the same way, and some people find this writing more understandable.

answered Feb 23, 2012 at 13:49

asaelr

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Dave · Accepted Answer · 2012-02-23 13:07:37Z

1

The signature you are looking for is:

void NodeToMIDI(unsigned node_pointer[][2], unsigned node_select, unsigned * MidiPacket);

which you would call

NodeToMIDI(node, /*something*/, /*something else*/);

answered Feb 23, 2012 at 13:07

Dave

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Jan Koester · Accepted Answer · 2012-02-23 13:13:49Z

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Some background information:

An array behaves a bit like a pointer. if you have char* name = {"John"}; you can access the characters with name[2] = h for example. Also name[0] (the first element) is the same like you use the pointer name directly. The pointer name refers to the adress of the first element in the array.

So an array of on array is... a pointer to a pointer. This looks like that what you described.

int ** a;

So this is an int pointer which referes to another int pointer which refers to a specific adress.

So you can take unsigned int** arrayOfArray as a function parameter. Then you can access the element like you know it from arrays.

arrayOfArray[0][2] = 42;

answered Feb 23, 2012 at 13:13

Jan Koester

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