Mixing
Why does the scoped enum support operator '<' by default?
Consider:
enum class Number {one, two};
if (Number::one < Number::two)
{}
My understanding is that the scoped enum needs to be cased into the underlying type or integer, and then it can be applied to operator < > ==. But it looks like the code snippet above can work without any explicit overloading operator <.
I don't see any descriptions in Enumeration declaration.
What does the C++ standard say about which operators are supported for a scoped enum by default?
c++ enums
edited Jan 28 at 19:25
Peter Mortensen
13.8k1987113
asked Jan 28 at 14:02
SSYSSY
31219
add a comment |
Consider:
enum class Number {one, two};
if (Number::one < Number::two)
{}
My understanding is that the scoped enum needs to be cased into the underlying type or integer, and then it can be applied to operator < > ==. But it looks like the code snippet above can work without any explicit overloading operator <.
I don't see any descriptions in Enumeration declaration.
What does the C++ standard say about which operators are supported for a scoped enum by default?
c++ enums
edited Jan 28 at 19:25
Peter Mortensen
13.8k1987113
asked Jan 28 at 14:02
SSYSSY
31219
The standard doesn't have a specific section for this. You need to check each expression and see if it support enumerations. It will say unscoped enumeration it if doesn't support scoped enumerations.
– NathanOliver
Jan 28 at 14:09
add a comment |
Consider:
enum class Number {one, two};
if (Number::one < Number::two)
{}
My understanding is that the scoped enum needs to be cased into the underlying type or integer, and then it can be applied to operator < > ==. But it looks like the code snippet above can work without any explicit overloading operator <.
I don't see any descriptions in Enumeration declaration.
What does the C++ standard say about which operators are supported for a scoped enum by default?
c++ enums
edited Jan 28 at 19:25
Peter Mortensen
13.8k1987113
asked Jan 28 at 14:02
SSYSSY
31219
Consider:
enum class Number {one, two};
if (Number::one < Number::two)
{}
My understanding is that the scoped enum needs to be cased into the underlying type or integer, and then it can be applied to operator < > ==. But it looks like the code snippet above can work without any explicit overloading operator <.
I don't see any descriptions in Enumeration declaration.
What does the C++ standard say about which operators are supported for a scoped enum by default?
c++ enums
c++ enums
edited Jan 28 at 19:25
Peter Mortensen
13.8k1987113
asked Jan 28 at 14:02
SSYSSY
31219
edited Jan 28 at 19:25
Peter Mortensen
13.8k1987113
asked Jan 28 at 14:02
SSYSSY
31219
edited Jan 28 at 19:25
Peter Mortensen
13.8k1987113
edited Jan 28 at 19:25
Peter Mortensen
13.8k1987113
edited Jan 28 at 19:25
Peter Mortensen
13.8k1987113
13.8k1987113
asked Jan 28 at 14:02
SSYSSY
31219
asked Jan 28 at 14:02
SSYSSY
31219
asked Jan 28 at 14:02
SSYSSY
31219
31219
The standard doesn't have a specific section for this. You need to check each expression and see if it support enumerations. It will say unscoped enumeration it if doesn't support scoped enumerations.
– NathanOliver
Jan 28 at 14:09
add a comment |
The standard doesn't have a specific section for this. You need to check each expression and see if it support enumerations. It will say unscoped enumeration it if doesn't support scoped enumerations.
– NathanOliver
Jan 28 at 14:09
The standard doesn't have a specific section for this. You need to check each expression and see if it support enumerations. It will say unscoped enumeration it if doesn't support scoped enumerations.
– NathanOliver
Jan 28 at 14:09
The standard doesn't have a specific section for this. You need to check each expression and see if it support enumerations. It will say unscoped enumeration it if doesn't support scoped enumerations.
– NathanOliver
Jan 28 at 14:09
add a comment |
2 Answers
2
active
oldest
votes
If you are referring to the "usual arithmetic conversions", then yes they are done when the arguments are arithmetic or enumeration types. It's just that there is a special bullet there for scoped enums:
[expr]
11 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,
which are defined as follows:
- 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.
So this case is covered here. Since the two operands are of the same scoped enum type, they are just checked to hold the specific relation in the fashion one would expect.
answered Jan 28 at 14:13
StoryTellerStoryTeller
104k12218281
add a comment |
My understanding is that scoped enum needs to be cased into underlying type or integer then it can be applied to operator < > ==.
Not when both of them are scoped enums. SomeScopedEnum < SomeInt is ill-formed, you're right in that case.
[expr.rel]p6:
If both operands (after conversions) are of arithmetic or enumeration type, each of the operators shall yield
trueif the specified relationship is true andfalseif it is false.
answered Jan 28 at 14:06
Rakete1111Rakete1111
35.2k1084119
Can I say operator < will use underlying integer to compare the scoped enum elements if they are same type?
– SSY
Jan 28 at 14:10
1
@SSY Yes, you can.
– Max Langhof
Jan 28 at 14:14
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you are referring to the "usual arithmetic conversions", then yes they are done when the arguments are arithmetic or enumeration types. It's just that there is a special bullet there for scoped enums:
[expr]
11 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,
which are defined as follows:
- 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.
So this case is covered here. Since the two operands are of the same scoped enum type, they are just checked to hold the specific relation in the fashion one would expect.
answered Jan 28 at 14:13
StoryTellerStoryTeller
104k12218281
add a comment |
If you are referring to the "usual arithmetic conversions", then yes they are done when the arguments are arithmetic or enumeration types. It's just that there is a special bullet there for scoped enums:
[expr]
11 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,
which are defined as follows:
- 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.
So this case is covered here. Since the two operands are of the same scoped enum type, they are just checked to hold the specific relation in the fashion one would expect.
answered Jan 28 at 14:13
StoryTellerStoryTeller
104k12218281
add a comment |
If you are referring to the "usual arithmetic conversions", then yes they are done when the arguments are arithmetic or enumeration types. It's just that there is a special bullet there for scoped enums:
[expr]
11 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,
which are defined as follows:
- 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.
So this case is covered here. Since the two operands are of the same scoped enum type, they are just checked to hold the specific relation in the fashion one would expect.
answered Jan 28 at 14:13
StoryTellerStoryTeller
104k12218281
If you are referring to the "usual arithmetic conversions", then yes they are done when the arguments are arithmetic or enumeration types. It's just that there is a special bullet there for scoped enums:
[expr]
11 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,
which are defined as follows:
- 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.
So this case is covered here. Since the two operands are of the same scoped enum type, they are just checked to hold the specific relation in the fashion one would expect.
answered Jan 28 at 14:13
StoryTellerStoryTeller
104k12218281
answered Jan 28 at 14:13
StoryTellerStoryTeller
104k12218281
answered Jan 28 at 14:13
StoryTellerStoryTeller
104k12218281
answered Jan 28 at 14:13
StoryTellerStoryTeller
104k12218281
104k12218281
add a comment |
add a comment |
My understanding is that scoped enum needs to be cased into underlying type or integer then it can be applied to operator < > ==.
Not when both of them are scoped enums. SomeScopedEnum < SomeInt is ill-formed, you're right in that case.
[expr.rel]p6:
If both operands (after conversions) are of arithmetic or enumeration type, each of the operators shall yield
trueif the specified relationship is true andfalseif it is false.
answered Jan 28 at 14:06
Rakete1111Rakete1111
35.2k1084119
Can I say operator < will use underlying integer to compare the scoped enum elements if they are same type?
– SSY
Jan 28 at 14:10
1
@SSY Yes, you can.
– Max Langhof
Jan 28 at 14:14
add a comment |
My understanding is that scoped enum needs to be cased into underlying type or integer then it can be applied to operator < > ==.
Not when both of them are scoped enums. SomeScopedEnum < SomeInt is ill-formed, you're right in that case.
[expr.rel]p6:
If both operands (after conversions) are of arithmetic or enumeration type, each of the operators shall yield
trueif the specified relationship is true andfalseif it is false.
answered Jan 28 at 14:06
Rakete1111Rakete1111
35.2k1084119
Can I say operator < will use underlying integer to compare the scoped enum elements if they are same type?
– SSY
Jan 28 at 14:10
1
@SSY Yes, you can.
– Max Langhof
Jan 28 at 14:14
add a comment |
My understanding is that scoped enum needs to be cased into underlying type or integer then it can be applied to operator < > ==.
Not when both of them are scoped enums. SomeScopedEnum < SomeInt is ill-formed, you're right in that case.
[expr.rel]p6:
If both operands (after conversions) are of arithmetic or enumeration type, each of the operators shall yield
trueif the specified relationship is true andfalseif it is false.
answered Jan 28 at 14:06
Rakete1111Rakete1111
35.2k1084119
My understanding is that scoped enum needs to be cased into underlying type or integer then it can be applied to operator < > ==.
Not when both of them are scoped enums. SomeScopedEnum < SomeInt is ill-formed, you're right in that case.
[expr.rel]p6:
If both operands (after conversions) are of arithmetic or enumeration type, each of the operators shall yield
trueif the specified relationship is true andfalseif it is false.
answered Jan 28 at 14:06
Rakete1111Rakete1111
35.2k1084119
answered Jan 28 at 14:06
Rakete1111Rakete1111
35.2k1084119
answered Jan 28 at 14:06
Rakete1111Rakete1111
35.2k1084119
answered Jan 28 at 14:06
Rakete1111Rakete1111
35.2k1084119
35.2k1084119
Can I say operator < will use underlying integer to compare the scoped enum elements if they are same type?
– SSY
Jan 28 at 14:10
1
@SSY Yes, you can.
– Max Langhof
Jan 28 at 14:14
add a comment |
Can I say operator < will use underlying integer to compare the scoped enum elements if they are same type?
– SSY
Jan 28 at 14:10
1
@SSY Yes, you can.
– Max Langhof
Jan 28 at 14:14
Can I say operator < will use underlying integer to compare the scoped enum elements if they are same type?
– SSY
Jan 28 at 14:10
Can I say operator < will use underlying integer to compare the scoped enum elements if they are same type?
– SSY
Jan 28 at 14:10
1
1
@SSY Yes, you can.
– Max Langhof
Jan 28 at 14:14
@SSY Yes, you can.
– Max Langhof
Jan 28 at 14:14
add a comment |
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The standard doesn't have a specific section for this. You need to check each expression and see if it support enumerations. It will say unscoped enumeration it if doesn't support scoped enumerations.
– NathanOliver
Jan 28 at 14:09