Thread: Multidimensional Arrays

It just needs to know the length of each row. It doesn't care how many rows you have, that's your job to keep track of. It's the same reason you aren't required to pass the length of a single array:

Code:

void foo( int bar[] );

It's actually because it changes arrays to a pointer to their first element's type when it passes them to a function.


Quzah.

The alternative way of saying it (which also reinforces the "it's a pointer to" idea) is
int func (int (*array)[20], int x);

There is no size (empty or otherwise), it's all the same to the compiler.

You can do this if you want.

Code:

// even if you specify a size, it doesn't make any difference.
int func (int array [5][20], int x);

int main ( ) {
  int a[100][20], b[200][20], c[20];
  func( a, 100 );
  func( b, 200 );
  func( &c, 1 );
}

Here's a simpler example
Saying
size_t strlen( char s[100] ); // not char* s
does not limit you to working out the length of strings which are only 100 characters long.