int *a[]
, when used as a function parameter (but not in normal declarations), is a pointer to a pointer, not a pointer to an array (in normal declarations, it is an array of pointers). A pointer to an array looks like this:
int (*aptr)[N]
Where N
is a particular positive integer (not a variable).
If you make your function a template, you can do it and you don't even need to pass the size of the array (because it is automatically deduced):
template<size_t SZ>
void generateArray(int (*aptr)[SZ])
{
for (size_t i=0; i<SZ; ++i)
(*aptr)[i] = rand() % 9;
}
int main()
{
int a[5];
generateArray(&a);
}
You could also take a reference:
template<size_t SZ>
void generateArray(int (&arr)[SZ])
{
for (size_t i=0; i<SZ; ++i)
arr[i] = rand() % 9;
}
int main()
{
int a[5];
generateArray(a);
}