Fix EnumeratorHandle compilation #5109
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Signed-off-by: Bili Dong <[email protected]>
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| @@ -630,6 +631,11 @@ const EnumeratorHandle<T> &EnumeratorHandle<T>::operator++() { | |||
| return *this; | |||
| } | |||
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| template <typename T> | |||
| bool EnumeratorHandle<T>::operator==(const EnumeratorHandle<T> &other) const { | |||
| return !(*this != other); | |||
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Sorry for the noise comment, but is this way of defining operator== common in C++?
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Usually the opposite (operator!= is implemented from operator==), however, here we already having operator!= with quite peculiar semantics.
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Usually the opposite (
operator!=is implemented fromoperator==),
Indeed, with C++20, you don't even need to do that -- if you define operator==, the compiler will synthesize operator!= for you
| @@ -630,6 +631,11 @@ const EnumeratorHandle<T> &EnumeratorHandle<T>::operator++() { | |||
| return *this; | |||
| } | |||
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| template <typename T> | |||
| bool EnumeratorHandle<T>::operator==(const EnumeratorHandle<T> &other) const { | |||
| return !(*this != other); | |||
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I guess this is as good as can be done, but the (existing) EnumeratorHandle::operator!= right after this looks suspect to me.
| @@ -630,6 +631,11 @@ const EnumeratorHandle<T> &EnumeratorHandle<T>::operator++() { | |||
| return *this; | |||
| } | |||
|
|
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| template <typename T> | |||
| bool EnumeratorHandle<T>::operator==(const EnumeratorHandle<T> &other) const { | |||
| return !(*this != other); | |||
There was a problem hiding this comment.
Usually the opposite (
operator!=is implemented fromoperator==),
Indeed, with C++20, you don't even need to do that -- if you define operator==, the compiler will synthesize operator!= for you
fruffy
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Fixes #5107.