sizeof(float)/sizeof(float[0])

[James R Twine]

That is an example of the decay I mentioned in my post - when you pass the array, it becomes a pointer in the called function.    The function that has the pointer has no "internal" or automatic way of knowing how much the data points to.  That is why whenever you pass an array of data to a function, you usually have to tell the function how much valid data is in the array (or have a special element in the array to indicate the end of valid data).    Peace!

-=- James

---

If you think it costs a lot to do it right, just wait until you find out how much it costs to do it wrong!
Avoid driving a vehicle taller than you and remember that Professional Driver on Closed Course does not mean your Dumb Ass on a Public Road!
DeleteFXPFiles & CheckFavorites (Please rate this post!)

[Alex Cutovoi]

ok, so how can I discover the array size? I don't want to use vectors....

[James R Twine]

You have to pass the size to the called function to let it know how much valid data exists.    Peace!

-=- James

---

If you think it costs a lot to do it right, just wait until you find out how much it costs to do it wrong!
Avoid driving a vehicle taller than you and remember that Professional Driver on Closed Course does not mean your Dumb Ass on a Public Road!
DeleteFXPFiles & CheckFavorites (Please rate this post!)

[Galatei]

Hi, Just modify Shape function by adding INT nCount after pointer to array of floats. Then in Shape function, calculate array size as nCount*sizeof(float). nCount should specify the size of the array. Regards

[Waldermort]

Depending on how often you are passing these arrays around and how frequently the size changes, you might want to declare a struct with the array and the size as members, then pass that to your functions.

typdef struct _double_array
{
double array[9];
int nCount;
} DOUBLE_ARRAY, *LPDOUBLE_ARRAY;

You can then pass the stuct 'DOUBLE_ARRAY' or a pointer to the struct 'LPDOUBLE_ARRAY'.

[Alex Cutovoi]

I made this: I've passed the size of the array(like you've said) and divided by float[0] and works fine

[Alex Cutovoi]

I've made this: primitive.Shape(GL_TRIANGLES, fVertexes, sizeof(fVertexes), 0,1,1); And in the function I'm divided the size of array by the size of first element in the array. Works fine, thanks for help

[Alex Cutovoi]

Thanks for the tip. It's a good idea

[Alex Cutovoi]

I made some modifications and works fine. Thanks for the support

[Waldermort]

No problem. If you look into the windows api's, you will find that many of the functions are declared like this, you either have to set up a struct or you get one back. My current project calls on the need of dealing with several arrays all relating to the same info, so instead of passing all the arrays and sizes to each function, I simply pass a pointer to a struct.

[Zac Howland]

If the array is allocated on the heap, sizeof behaves differently than it does if it was on the stack.

int i = 0;

float stackArray[5];
i = sizeof(stackArray) / sizeof(stackArray[0]);	// i is 5
// sizeof(stackArray) will return (5 * sizeof(float))
// sizeof(stackArray[0]) will return sizeof(float)

float* heapArray = new float[5];
i = sizeof(heapArray) / sizeof(heapArray[0]);	// i is 1
// sizeof(heapArray) will return sizeof(float*)
// sizeof(heapArray[0]) will return sizeof(float)

// Interesting Note (forget bounds errors here)
int myFunction(float* myArray)
{
	return sizeof(myArray) / sizeof(myArray[0]);
}
// above function will return heap results regardless of where
// the float array is actually allocated (that is, always 1)

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week Zac