Many binary operators that expect operands ofarithmetic orenumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called theusual arithmetic conversions.

[edit]Definition

Usual arithmetic conversions are defined as follows:

[edit]Stage 1

Applieslvalue-to-rvalue conversion to both operands, the resulting prvalues are used in place of the original operands for the remaining process.

[edit]Stage 2

[edit]Stage 3

[edit]Stage 4

• If either operand is offloating-point type, the following rules are applied:

• If both operands have the same type, no further conversion will be performed.
• Otherwise, if one of the operands is of a non-floating-point type, that operand is converted to the type of the other operand.
• Otherwise, if thefloating-point conversion ranks of the types of the operands areordered but(since C++23) not equal, then the operand of the type with the lesser floating-point conversion rank is converted to the type of the other operand.

• Otherwise, both operands are of integer types, proceed to the next stage.

[edit]Stage 5

Both operands are converted to a common typeC. Given the typesT1 andT2 as the promoted type (under the rules of integral promotions) of the operands, the following rules are applied to determineC:

• IfT1 andT2 are the same type,C is that type.
• Otherwise, ifT1 andT2 are both signed integer types or both unsigned integer types,C is the type of greaterinteger conversion rank.
• Otherwise, one type betweenT1 andT2 is an signed integer typeS, the other type is an unsigned integer typeU. Apply the following rules:

• If the integer conversion rank ofU is greater than or equal to the integer conversion rank ofS,C isU.
• Otherwise, ifS can represent all of the values ofU,C isS.
• Otherwise,C is the unsigned integer type corresponding toS.

[edit]Integer conversion rank

Everyinteger type has aninteger conversion rank defined as follows:

• No two signed integer types other thanchar andsignedchar (ifchar is signed) have the same rank, even if they have the same representation.
• The rank of a signed integer type is greater than the rank of any signed integer type with a smaller width.
• The ranks of the following integer types decrease in order:

• long
• int
• short
• signedchar

• The rank of any unsigned integer type equals the rank of the corresponding signed integer type.

• The rank ofbool is less than the rank of all standard integer types.
• The ranks of encoding character types (char,char8_t(since C++20),char16_t,char32_t,(since C++11) andwchar_t) equal the ranks of theirunderlying types, which means:

• The rank ofchar equals the rank ofsignedchar andunsignedchar.

• The rank ofwchar_t equals the rank of its implementation-defined underlying type.

• For all integer typesT1,T2, andT3, ifT1 has greater rank thanT2 andT2 has greater rank thanT3, thenT1 has greater rank thanT3.

The integer conversion rank is also used in the definition ofintegral promotion.

[edit]Floating-point conversion rank and subrank

[edit]Floating-point conversion rank

Everyfloating-point type has afloating-point conversion rank defined as follows:

• The ranks of the standard floating-point types decrease in order:
  • longdouble
  • double
  • float

[edit]Usage

The floating-point conversion rank and subrank are also used to

• determine whether a conversion between different floating-point typescan be implicit or is anarrowing conversion,
• distinguish the conversion sequences in overload resolution,

• determine whetherstd::complex'sconverting constructor is explicit, or
• determine the common floating-point type if the arguments of different floating-point types are passed tocommon orspecial math functions.

[edit]Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

1. ↑Before the resolution,unsignedchar is promoted toint at the beginning of stage 5, then it is converted tounsignedint. However, the latter conversion is narrowing, which makes the three-way comparison ill-formed.
2. ↑After the resolution, the common type is stillunsignedint. The difference is thatunsignedchar is directly converted tounsignedint without the intermediate integral promotion. The conversion is not narrowing and hence the three-way comparison is well-formed.

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