You indexing in that for() loop isn't right.
You need to add 2 to i on every pass or
increment it in the second [i+1] access,
and assuming "size" is the length of "array"
you then shouldn't divide by 2 either...
Or you can just throw away the top 16 bits of
each rand16 to make it easier:
uint8_t array[size];
for (i=0; i < size; i++)
{
aux = call Random.rand16();
array[i] = aux & 0xff;
}
I would do some study of the actual values
produced by rand16() to see how random each
byte is as well. I've seen (very) simple
pseudo-random generators that have large
jumps between values that are not as random
as one might like.
MS
Ruben Rios wrote:
> Hello everyone,
>
> I have seen the Random interface generates random numbers of 16 or 32
> bits long. How can I generate a 8 bit random numbers array?
>
> void generateRandomArray(uint8_t * array, uint8_t size)
> {
> uint8_t i;
> uint16_t aux;
> uint8_t *p = (uint8_t *)&aux;
>
> for (i=0; i < size / 2; i++)
> {
> aux = call Random.rand16();
> array[i] = *p; //Store the MSB
> array[i+1] = *(p+1); //Store the LSB
> }
> }
>
> I am not very good at pointers that's why I don't know if this works
> fine. By the way, I guess the array contents are modified within the
> function but these are also valid outside, right?
>
> Thanks in advance
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