if we can pass array size in square brackets ?when function is declared or array is pass as an argument

Answer 1

You can pass an array in like this:

void func(int arr[])
{
//code
}

int main()
{
int arr[5];
func(arr);
}

(pass by reference if you need to)

or if you only want to pass the array's size:

void func(int b)
{
//code
}

int main()
{
const int a = 5;
int arr[5];
func(x);
}

Answer 2

no u can't

Answer 3

u can only write the array name as a parameter and it's size as another parameter inside void functions. But don't forget to separate the two parameters with a comma

Answer 4

You can use sizeof operator. sizeof(arr) should return the same number as you see in the square bracket.

Answer 5

that's not my question i wanna know if i can pass array size in( void func )inside the square brackets

Answer 6

ok thanks

Answer 7

That's not possible.
Say you have a function
func()
that takes an array as an argument, in that case you will only pass the name of the array, without the square brackets, to that function:
int arr[3]={2,7,4};
func(arr);
if you were to pass the name WITH the square brackets, you get an error, and if you were to pass arr[2] for example, it would be resolved to the CONTENT of arr[2].
Finally, for an array with the length x, the highest index would be x-1, so if arr had a length of 3, the indices woul be 0 1 2 , thus, arr[3] would point outside the range of the array.