Passing a 2D array as a pointer to a pointer

Say I have a function like this.

int Func(float **fppArg);

and I have a variable defined thus.

float faa2DArray[3][3];

How would I pass faa2DArray to Func()?

Many thanks in advance,
Peter.

PeterOut <MajorSetback@excite.comwrites:
You can't, at least not directly. You can write:

float *tmp[] = { faa2DArray[0], faa2DArray[1], faa2DArray[2] };
Func(tmp);

or even:

Func((float *[]){faa2DArray[0], faa2DArray[1], faa2DArray[2]});

if you don't mind straying into C99. This is more universal:

float *tmp[3];
tmp[0] = faa2DArray[0];
tmp[1] = faa2DArray[1];
tmp[2] = faa2DArray[2];
Func(tmp);

However, the fact that you need these gymnastics suggests that
something has gone wrong. Can't you start with the right shape of
array in the first place, or change Func to take the array you have?

--
Ben.

PeterOut wrote:
You wouldn't. You have to change the declaration of either fppArg or
faa2DArray.


One way would be:

int Func(float fppArg[][3]);



Brian

On 27 Aug, 23:53, PeterOut <MajorSetb...@excite.comwrote:http://c-faq.com/

FAQ 6.18 "My compiler complained when I passed a two-dimensional
array to a function expecting a pointer to a pointer. "

then read all of section 6. Then read the rest of the FAQ.

--
Nick Keighley

"Resistance is futile. Read the C-faq."
-- James Hu (c.l.c.)