22 General utilities library [utilities]

22.5 Optional objects [optional]

22.5.8 Comparison with T [optional.comp.with.t]

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template<class T, class U> constexpr bool operator==(const optional<T>& x, const U& v);
1
#
Constraints: U is not a specialization of optional.
The expression *x == v is well-formed and its result is convertible to bool.
[Note 1:  — end note]
2
#
Effects: Equivalent to: return x.has_value() ? *x == v : false;
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template<class T, class U> constexpr bool operator==(const T& v, const optional<U>& x);
3
#
Constraints: T is not a specialization of optional.
The expression v == *x is well-formed and its result is convertible to bool.
4
#
Effects: Equivalent to: return x.has_value() ? v == *x : false;
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template<class T, class U> constexpr bool operator!=(const optional<T>& x, const U& v);
5
#
Constraints: U is not a specialization of optional.
The expression *x != v is well-formed and its result is convertible to bool.
6
#
Effects: Equivalent to: return x.has_value() ? *x != v : true;
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template<class T, class U> constexpr bool operator!=(const T& v, const optional<U>& x);
7
#
Constraints: T is not a specialization of optional.
The expression v != *x is well-formed and its result is convertible to bool.
8
#
Effects: Equivalent to: return x.has_value() ? v != *x : true;
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template<class T, class U> constexpr bool operator<(const optional<T>& x, const U& v);
9
#
Constraints: U is not a specialization of optional.
The expression *x < v is well-formed and its result is convertible to bool.
10
#
Effects: Equivalent to: return x.has_value() ? *x < v : true;
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template<class T, class U> constexpr bool operator<(const T& v, const optional<U>& x);
11
#
Constraints: T is not a specialization of optional.
The expression v < *x is well-formed and its result is convertible to bool.
12
#
Effects: Equivalent to: return x.has_value() ? v < *x : false;
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template<class T, class U> constexpr bool operator>(const optional<T>& x, const U& v);
13
#
Constraints: U is not a specialization of optional.
The expression *x > v is well-formed and its result is convertible to bool.
14
#
Effects: Equivalent to: return x.has_value() ? *x > v : false;
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template<class T, class U> constexpr bool operator>(const T& v, const optional<U>& x);
15
#
Constraints: T is not a specialization of optional.
The expression v > *x is well-formed and its result is convertible to bool.
16
#
Effects: Equivalent to: return x.has_value() ? v > *x : true;
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template<class T, class U> constexpr bool operator<=(const optional<T>& x, const U& v);
17
#
Constraints: U is not a specialization of optional.
The expression *x <= v is well-formed and its result is convertible to bool.
18
#
Effects: Equivalent to: return x.has_value() ? *x <= v : true;
🔗
template<class T, class U> constexpr bool operator<=(const T& v, const optional<U>& x);
19
#
Constraints: T is not a specialization of optional.
The expression v <= *x is well-formed and its result is convertible to bool.
20
#
Effects: Equivalent to: return x.has_value() ? v <= *x : false;
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template<class T, class U> constexpr bool operator>=(const optional<T>& x, const U& v);
21
#
Constraints: U is not a specialization of optional.
The expression *x >= v is well-formed and its result is convertible to bool.
22
#
Effects: Equivalent to: return x.has_value() ? *x >= v : false;
🔗
template<class T, class U> constexpr bool operator>=(const T& v, const optional<U>& x);
23
#
Constraints: T is not a specialization of optional.
The expression v >= *x is well-formed and its result is convertible to bool.
24
#
Effects: Equivalent to: return x.has_value() ? v >= *x : true;
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template<class T, class U> requires (!is-derived-from-optional<U>) && three_way_comparable_with<T, U> constexpr compare_three_way_result_t<T, U> operator<=>(const optional<T>& x, const U& v);
25
#
Effects: Equivalent to: return x.has_value() ? *x <=> v : strong_ordering​::​less;
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