Pointers and their length

[AmitM9S6]

I'm trying to write a simple C program, though for some reason, I'm having a problem getting the array's size

I know I need to use sizeof, but it doesnt work inside the function :

void printArr(int arr [])
{
	for (int i = 0 ; i < n ; i++)
	{
		printf("%d ", arr[i]);
	}
	printf("\n");
}

int main()
{
	int c[6] = {1, 3, 2, 5, 0, 7};
	int n = sizeof(c)/sizeof(int);
	printArr(c);

	getchar();
	getchar();
}

in the main, n will be 6, though inside the function printArr, n will be 1. why is that?

[Peter87]

Line 1 is equivalent to

void printArr(int* arr)

so the problem is that sizeof(arr) will give you the size of the pointer (usually 4 or 8 bytes).

There is no way you can know the array size from just the pointer. You will have to pass the size as an additional argument to the function.

[giblit]

in the main, n will be 6, though inside the function printArr, n will be 1. why is that?

It should be undefined and not compile since it is not passed to the function.

[AmitM9S6]

Thank you guys! I thought adding a size paramter is a bad programming, because you could check it urself at the function!

[Duthomhas]

For more on the size of arrays:
http://www.cplusplus.com/faq/sequences/arrays/sizeof-array/