For no particular reason, arrays cannot be assigned to one another. Use std::copy instead:
#include <algorithm>
// ...
int a[8] = {2, 3, 5, 7, 11, 13, 17, 19};
int b[8];
std::copy(a + 0, a + 8, b);
This is more flexible than what true array assignment could provide because it is possible to copy slices of larger arrays into smaller arrays.
std::copy is usually specialized for primitive types to give maximum performance. It is unlikely that std::memcpy performs better. If in doubt, measure.
Although you cannot assign arrays directly, you can assign structs and classes which contain array members. That is because array members are copied memberwise by the assignment operator which is provided as a default by the compiler. If you define the assignment operator manually for your own struct or class types, you must fall back to manual copying for the array members.
Arrays cannot be passed by value. You can either pass them by pointer or by reference.
Since arrays themselves cannot be passed by value, usually a pointer to their first element is passed by value instead. This is often called "pass by pointer". Since the size of the array is not retrievable via that pointer, you have to pass a second parameter indicating the size of the array (the classic C solution) or a second pointer pointing after the last element of the array (the C++ iterator solution):
#include <numeric>
#include <cstddef>
int sum(const int* p, std::size_t n)
{
return std::accumulate(p, p + n, 0);
}
int sum(const int* p, const int* q)
{
return std::accumulate(p, q, 0);
}
As a syntactic alternative, you can also declare parameters as T p[], and it means the exact same thing as T* p in the context of parameter lists only:
int sum(const int p[], std::size_t n)
{
return std::accumulate(p, p + n, 0);
}
You can think of the compiler as rewriting T p[] to T *p in the context of parameter lists only. This special rule is partly responsible for the whole confusion about arrays and pointers. In every other context, declaring something as an array or as a pointer makes a huge difference.
Unfortunately, you can also provide a size in an array parameter which is silently ignored by the compiler. That is, the following three signatures are exactly equivalent, as indicated by the compiler errors:
int sum(const int* p, std::size_t n)
// error: redefinition of 'int sum(const int*, size_t)'
int sum(const int p[], std::size_t n)
// error: redefinition of 'int sum(const int*, size_t)'
int sum(const int p[8], std::size_t n) // the 8 has no meaning here
Arrays can also be passed by reference:
int sum(const int (&a)[8])
{
return std::accumulate(a + 0, a + 8, 0);
}
In this case, the array size is significant. Since writing a function that only accepts arrays of exactly 8 elements is of little use, programmers usually write such functions as templates:
template <std::size_t n>
int sum(const int (&a)[n])
{
return std::accumulate(a + 0, a + n, 0);
}
Note that you can only call such a function template with an actual array of integers, not with a pointer to an integer. The size of the array is automatically inferred, and for every size n, a different function is instantiated from the template. You can also write quite useful function templates that abstract from both the element type and from the size.