C++ – using const reference to prolong a member of a temporary, ok or UB?

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Your code should be well-formed, because for temporaries

(emphasis mine)

Whenever a reference is bound to a temporary or to a subobject thereof, the lifetime of the temporary is extended to match the lifetime of the reference

Given A().b[4], b[4] is the subobject of b and the data member b is the subobject of the temproray A(), whose lifetime should be extended.

LIVE on clang10 with -O2
LIVE on gcc10 with -O2

BTW: This seems to be a gcc’s bug which has been fixed.

From the standard, [class.temporary]/6

The third context is when a reference is bound to a temporary object.36 The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference if the glvalue to which the reference is bound was obtained through one of the following:

…

[ Example:

template<typename T> using id = T;

int i = 1;
int&& a = id<int[3]>{1, 2, 3}[i];          // temporary array has same lifetime as a
const int& b = static_cast<const int&>(0); // temporary int has same lifetime as b
int&& c = cond ? id<int[3]>{1, 2, 3}[i] : static_cast<int&&>(0);
                                           // exactly one of the two temporaries is lifetime-extended

— end example ]

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