C++ Space occupied by a reference


Example

A reference is not an object, and unlike an object, it is not guaranteed to occupy some contiguous bytes of memory. The standard leaves it unspecified whether a reference requires any storage at all. A number of features of the language conspire to make it impossible to portably examine any storage the reference might occupy:

  • If sizeof is applied to a reference, it returns the size of the referenced type, thereby giving no information about whether the reference occupies any storage.
  • Arrays of references are illegal, so it is not possible to examine the addresses of two consecutive elements of a hypothetical reference of arrays in order to determine the size of a reference.
  • If the address of a reference is taken, the result is the address of the referent, so we cannot get a pointer to the reference itself.
  • If a class has a reference member, attempting to extract the address of that member using offsetof yields undefined behavior since such a class is not a standard-layout class.
  • If a class has a reference member, the class is no longer standard layout, so attempts to access any data used to store the reference results in undefined or unspecified behavior.

In practice, in some cases a reference variable may be implemented similarly to a pointer variable and hence occupy the same amount of storage as a pointer, while in other cases a reference may occupy no space at all since it can be optimized out. For example, in:

void f() {
    int x;
    int& r = x;
    // do something with r
}

the compiler is free to simply treat r as an alias for x and replace all occurrences of r in the rest of the function f with x, and not allocate any storage to hold r.