Thread: arrays and pointers

given
double ar[20];

and a function call
function(ar,size);

is this equivalent to
function(&ar[0],size)
?????

whats the difference between
function(ar+1,size)
and
function(&ar[1],size)
????

Presumably, your function declaration would look like this:

void function(double* ar, int size);

In this case, the calls:


function(ar, 20);

and

function (&ar[0], 20);


are equivelent. In both cases you are passing a pointer to the first element in an array of 20 doubles. In the second case, you have de-referenced the first element to a double, and then re-referenced it to a pointer with the & character.

You can use this technique to pass a pointer to the second character (as your example):

function(&ar[1], 19);

Notice I have decremented the size value, as there are only 19 elements in the array pointed to by &ar[1].

The 'function(ar+1,size)' looks incorrect. Remember ar is an 'address' of an array of doubles, not a double itself.

The 'function(ar+1,size)' looks incorrect. Remember ar is an 'address' of an array of doubles, not a double itself.

function(ar+1, size-1);
is the same as
function(&ar[1], 19);

the 'ar+1' increments the address stored in ar by the size of 1 double.

OK, I take it back. &ar[1] & ar + 1 are equivalent in that case. Never used that notation before.

Bravo, Larry! Thank you for pointing that out. Pointer arithmetic is very important in programming C/C++.