For 2D array, we can pass the array to other function like:
I' m wondering how we can pass a 3D array...Code:int array[5][10]; ... func((int *)array, 5, 10); ... ... int func(int *array, int x, int y) { *(array + y * 0 + 8) = 1000; /*change array[0][8]*/ *(array + y * 1 + 8) = 2000; /*change array[1][8]*/ }
> For 2D array, we can pass the array to other function like:
Since your attempt at 2D is wrong, 3D will be hopeless.
If you have
int array[5][10];
The prototype for a function is
void func ( int array[5][10] );
A call to this function would be
func ( array );
And the implementation would be
Notice how simply copy/pasting the array into the function parameter is all you need to do.Code:void func ( int array[5][10] ) { array[0][0] = 0; }
Also notice the complete lack of any * or & operators anywhere in the code.
Pedantically, you can reduce
void func ( int array[5][10] );
to
void func ( int array[][10] );
since C doesn't need to know the size of the first dimension of the array.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
What if we don't know exactly the size of array we gonna pass to func()?
How can we create a func() which can accept any size of array?
http://www.eskimo.com/~scs/C-faq/q6.19.html
int a[5][10];
int b[100][10];
Can both be passed to void func ( int array[][10] );
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
