sizeof

I have a function that relies on the sizeof statement to get the size of a constant array of unsigned bytes. Everytime I use the function the sizeof statement returns 4 regardless of the size of my array.

Here's the code that doesn't seem to work. The array argument is defined as a const u8 array[].

u8  c;
   
for(c = 0;c <= sizeof(array);++c){
    //do something here
}

1. Use fprintf ("fast printf") instead of printf.
2. ++i is faster than both i++ and i = i + 1.
3. void main(void) is faster than int main(void) or int main(int, char **) since no value needs to be returned to the OS.
4. Swapping with exclusive-or (a^=b^=a^=b swaps a and b) is faster than using a temporary. This works for all types (including structures), but not on all compilers. Some compilers may also give you a harmless warning.
5. Static storage duration objects are faster than automatic storage duration objects because the CPU doesn't have to set aside storage on the stack every time a function is called. Make your loop indexes global so that you can use them everywhere:
int i;
void func(void) { for (i = 0; i < 10; i++) ; /* ... */ }
void func2(void) { for (i = 0; i < 20; i++) ; /* ... */ }
/* ... */
6. Compilers often give more memory to arrays than you asked for. Here's how to check how big an array actually is (memset returns a null pointer if the size you passed to it is bigger than the size of the array you passed to it):
int arr[256];
size_t realsize;
for (realsize = 0; realsize <= SIZE_MAX; ++realsize)
        if (!memset(arr, 0, realsize)) break;
/* now you know that arr actually has realsize / sizeof (int) elements */
If you combine this with #5, your program will be faster in the long run (but this usually doesn't work for short programs).