Usual arithmetic conversions - cppreference.com (original) (raw)
Many binary operators that expect operands of arithmetic or enumeration 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 the usual arithmetic conversions.
Contents
- 1 Definition
- 2 Integer conversion rank
- 3 Floating-point conversion rank and subrank
- 4 Defect reports
[edit] Definition
Usual arithmetic conversions are defined as follows:
[edit] Stage 1
Applies lvalue-to-rvalue conversion to both operands, the resulting prvalues are used in place of the original operands for the remaining process.
[edit] Stage 2
| If either operand is of scoped enumeration type, no conversions are performed; if the other operand does not have the same type, the expression is ill-formed. Otherwise, proceed to the next stage. | (since C++11) |
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[edit] Stage 3
| If either operand is of enumeration type, and the other operand is of a different enumeration type or a floating-point type, the expression is ill-formed. Otherwise, proceed to the next stage. | (since C++26) |
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[edit] Stage 4
If either operand is of floating-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 the floating-point conversion ranks of the types of the operands are ordered 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, if the floating-point conversion ranks of the types of the operands are equal, then the operand with the lesser floating-point conversion subrank is converted to the type of the other operand. Otherwise, the expression is ill-formed. | (since C++23) |
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- Otherwise, both operands are of integer types, proceed to the next stage.
[edit] Stage 5
Both operands are converted to a common type C. Given the types T1 and T2 as the promoted type (under the rules of integral promotions) of the operands, the following rules are applied to determine C:
If
T1andT2are the same type,Cis that type.Otherwise, if
T1andT2are both signed integer types or both unsigned integer types,Cis the type of greater integer conversion rank.Otherwise, one type between
T1andT2is an signed integer typeS, the other type is an unsigned integer typeU. Apply the following rules:If the integer conversion rank of
Uis greater than or equal to the integer conversion rank ofS,CisU.Otherwise, if
Scan represent all of the values ofU,CisS.Otherwise,
Cis the unsigned integer type corresponding toS.
| If one operand is of enumeration type and the other operand is of a different enumeration type or a floating-point type, this behavior is deprecated. | (since C++20)(until C++26) |
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[edit] Integer conversion rank
Every integer type has an integer conversion rank defined as follows:
No two signed integer types other than char and signed char (if char 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
signed char
The rank of any unsigned integer type equals the rank of the corresponding signed integer type.
| The rank of any standard integer type is greater than the rank of any extended integer type with the same width. | (since C++11) |
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The rank of bool 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) and wchar_t) equal the ranks of their underlying types, which means:
The rank of char equals the rank of signed char and unsigned char.
| The rank of char8_t equals the rank of unsigned char. | (since C++20) |
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| The rank of char16_t equals the rank of std::uint_least16_t. The rank of char32_t equals the rank of std::uint_least32_t. | (since C++11) |
- The rank of wchar_t equals the rank of its implementation-defined underlying type.
| The rank of any extended signed integer type relative to another extended signed integer type with the same width is implementation-defined, but still subject to the other rules for determining the integer conversion rank. | (since C++11) |
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- For all integer types
T1,T2, andT3, ifT1has greater rank thanT2andT2has greater rank thanT3, thenT1has greater rank thanT3.
The integer conversion rank is also used in the definition of integral promotion.
[edit] Floating-point conversion rank and subrank
[edit] Floating-point conversion rank
Every floating-point type has a floating-point conversion rank defined as follows:
- The ranks of the standard floating-point types decrease in order:
- long double
- double
- float
| The rank of a floating-point type T is greater than the rank of any floating-point type whose set of values is a proper subset of the set of values of T. Two extended floating-point types with the same set of values have equal ranks. An extended floating-point type with the same set of values as exactly one cv-unqualified standard floating-point type has a rank equal to the rank of that standard floating-point type. An extended floating-point type with the same set of values as more than one cv-unqualified standard floating-point type has a rank equal to the rank of double. | (since C++23) |
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| Floating-point conversion subrank Floating-point types that have equal floating-point conversion ranks are ordered by floating-point conversion subrank. The subrank forms a total order among types with equal ranks.The types std::float16_t, std::float32_t, std::float64_t, and std::float128_t (fixed width floating-point types) have a greater conversion subrank than any standard floating-point type with equal conversion rank. Otherwise, the conversion subrank order is implementation-defined. | (since C++23) |
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[edit] Usage
The floating-point conversion rank and subrank are also used to
- determine whether a conversion between different floating-point types can be implicit or is a narrowing conversion,
- distinguish the conversion sequences in overload resolution,
| determine the actual type extracted by std::num_get::get() for the extraction of an extended floating-point type using std::basic_istream::operator>>, determinte the actual type inserted by std::num_put::put() for the insertion of an extended floating-point type using std::basic_ostream::operator<<, | (since C++23) |
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- determine whether std::complex's converting constructor is explicit, or
- determine the common floating-point type if the arguments of different floating-point types are passed to common or special math functions.
[edit] Defect reports
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
| DR | Applied to | Behavior as published | Correct behavior |
|---|---|---|---|
| CWG 1642 | C++98 | usual arithmetic conversions might involve lvalues | applies lvalue-to-rvalue conversions first |
| CWG 2528 | C++20 | the three-way comparison between unsigned charand unsigned int is ill-formed becauseof the intermediate integral promotion[1] | determines the common type basedon the promoted types, withoutactually promoting the operands[2] |
| CWG 2892 | C++98 | when both operands are of the samefloating-point type, the meaning of “nofurther conversion is needed” was unclear | changed to “no furtherconversion will be performed” |
- ↑ Before the resolution, unsigned char is promoted to int at the beginning of stage 5, then it is converted to unsigned int. However, the latter conversion is narrowing, which makes the three-way comparison ill-formed.
- ↑ After the resolution, the common type is still unsigned int. The difference is that unsigned char is directly converted to unsigned int without the intermediate integral promotion. The conversion is not narrowing and hence the three-way comparison is well-formed.