Re: How to know the size of array


manochavishal@xxxxxxxxx wrote:
void foo(int * array)



void foo(int *array, size_t len)


{


Now the array size is in `len`.

Thanx for the quick reply.

I think i read somewhere that i can get the size by:

(sizeof(array) ) / (sizeof(element))

in this case it will be

(sizeof(array)) / (sizeof(int))


Given the following:

int foo[10];

sizeof foo will return the total number of bytes taken up by the whole
array, and sizeof foo[0] will return the number of bytes used by a
single array element, so yes, the expression

sizeof foo / sizeof foo[0]

will give you the number of elements (10) in the array.

Unfortunately, this won't help you if you pass the array to a function.
In most contexts*, the type of the array identifier "decays" into a
pointer to the base type, and its value is set to the address of the
first element in the array. IOW, when you write

bar(foo);

what actually gets passed to bar is a pointer type object, not an array
type object:

void bar(int *arr)
{
...
}

and sizeof arr will return the number of bytes used by the pointer,
*not* the array it points to, so the sizeof arr / sizeof arr[0] trick
won't work.

* The only times the array identifier doesn't decay into a pointer is
when it's an operand of the sizeof and unary & operators.

I am not sure about this. Can we get the sizeof array to return the
size of all its elements.
Can sizeof find where the array is getting finished or it will just
give me the size of element.

Is there no way to get the size of array without passing the its
length.

Not really, no.


Cheers
Vishal

.

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