Defined in header
common_reference_with<T, U> specifies that two types
U share a common reference type (as computed by std::common_reference_t) to which both can be converted.
 Semantic requirements
T and U model
common_reference_with<T, U> only if, given equality-preserving expressions
u2 such that decltype((t1)) and decltype((t2)) are both
T and decltype((u1)) and decltype((u2)) are both
- std::common_reference_t<T, U>(t1) equals std::common_reference_t<T, U>(t2) if and only if
- std::common_reference_t<T, U>(u1) equals std::common_reference_t<T, U>(u2) if and only if
In other words, the conversion to the common reference type must preserve equality.
 Equality preservation
An expression is equality preserving if it results in equal outputs given equal inputs.
- The inputs to an expression consist of its operands.
- The outputs of an expression consist of its result and all operands modified by the expression (if any).
In specification of standard concepts, operands are defined as the largest subexpressions that include only:
The cv-qualification and value category of each operand is determined by assuming that each template type parameter denotes a cv-unqualified complete non-array object type.
Every expression required to be equality preserving is further required to be stable: two evaluations of such an expression with the same input objects must have equal outputs absent any explicit intervening modification of those input objects.
 See also
| determines the common reference type of a group of types |
| specifies that two types share a common type |
| determines the common type of a group of types |