A pointer can be re-assigned:
int x = 5;
int y = 6;
int *p;
p = &x;
p = &y;
*p = 10;
assert(x == 5);
assert(y == 10);
A reference cannot be re-bound, and must be bound at initialization:
int x = 5;
int y = 6;
int &q; // error
int &r = x;
A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary &
operator and a certain amount of space that can be measured with the sizeof
operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.
int x = 0;
int &r = x;
int *p = &x;
int *p2 = &r;
assert(p == p2); // &x == &r
assert(&p != &p2);
You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.
int x = 0;
int y = 0;
int *p = &x;
int *q = &y;
int **pp = &p;
**pp = 2;
pp = &q; // *pp is now q
**pp = 4;
assert(y == 4);
assert(x == 2);
A pointer can be assigned nullptr
, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference to nullptr
, but this is undefined and will not behave consistently.
/* the code below is undefined; your compiler may optimise it
* differently, emit warnings, or outright refuse to compile it */
int &r = *static_cast<int *>(nullptr);
// prints "null" under GCC 10
std::cout
<< (&r != nullptr
? "not null" : "null")
<< std::endl;
bool f(int &r) { return &r != nullptr; }
// prints "not null" under GCC 10
std::cout
<< (f(*static_cast<int *>(nullptr))
? "not null" : "null")
<< std::endl;
You can, however, have a reference to a pointer whose value is nullptr
.
Pointers can iterate over an array; you can use ++
to go to the next item that a pointer is pointing to, and + 4
to go to the 5th element. This is no matter what size the object is that the pointer points to.
A pointer needs to be dereferenced with *
to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses ->
to access its members whereas a reference uses a .
.
References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
const int &x = int(12); // legal C++
int *y = &int(12); // illegal to take the address of a temporary.
This makes const &
more convenient to use in argument lists and so forth.
Best Answer
Introduction
For a technical overview - skip to this answer.
For common cases where copy elision occurs - skip to this answer.
Copy elision is an optimization implemented by most compilers to prevent extra (potentially expensive) copies in certain situations. It makes returning by value or pass-by-value feasible in practice (restrictions apply).
It's the only form of optimization that elides (ha!) the as-if rule - copy elision can be applied even if copying/moving the object has side-effects.
The following example taken from Wikipedia:
Depending on the compiler & settings, the following outputs are all valid:
This also means fewer objects can be created, so you also can't rely on a specific number of destructors being called. You shouldn't have critical logic inside copy/move-constructors or destructors, as you can't rely on them being called.
If a call to a copy or move constructor is elided, that constructor must still exist and must be accessible. This ensures that copy elision does not allow copying objects which are not normally copyable, e.g. because they have a private or deleted copy/move constructor.
C++17: As of C++17, Copy Elision is guaranteed when an object is returned directly: