Specify copy assignment operator for vec swizzles #678
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Even though we specify an assignment operator for `__writeable_swizzle__` like this: ``` template</*unspecified*/> const __writeable_swizzle__& operator=(const __writeable_swizzle__</*unspecified*/>& rhs) const ``` The compiler will not select this operator when the left-hand-side has the same type as the right-hand-side. (Even though the template constraints would allow such a selection.) Instead, the compiler will generate the default copy assignment operator and call that. Since the default copy assignment operator would not be correct (performing a shallow copy), the `__writeable_swizzle__` class must provide a user defined copy assignment operator. The copy assignment operator for `__const_swizzle__` was already deleted in the "constructors" section. It seemed better to move this to the "member functions" part of the specification, next to the specification for the copy assignment operator for `__writeable_swizzle__`.
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Even though we specify an assignment operator for
__writeable_swizzle__like this:The compiler will not select this operator when the left-hand-side has the same type as the right-hand-side. (Even though the template constraints would allow such a selection.) Instead, the compiler will generate the default copy assignment operator and call that.
Since the default copy assignment operator would not be correct (performing a shallow copy), the
__writeable_swizzle__class must provide a user defined copy assignment operator.The copy assignment operator for
__const_swizzle__was already deleted in the "constructors" section. It seemed better to move this to the "member functions" part of the specification, next to the specification for the copy assignment operator for__writeable_swizzle__.