std - C++ default allocator - what should happen if the size doesn't equal the size passed to the invocation of allocate? -

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20.6.9:

void deallocate(pointer p, size_type n);
  • requires: p shall pointer value obtained allocate(). n shall equal value passed first argument invocation of allocate returned p.
  • effects: deallocates storage referenced p.
  • remarks: uses ::operator delete(void*) (18.6.1), unspecified when function called.

what should happen if ndoesn't equal value passed first agrgument invocation of allocate returned p? not deallocate? throw std::bad_alloc? ...

edit: meant "what should happen" was: okay throw or assert in custom implementation?

as usual in c++ standard, when nothing stated explicitly, violating requirements leads undefined behavior. shall means at times must, it's requirement, not option in c++ standard.

for example here's msdn says:

the pointer _ptr must have been returned earlier call allocate allocator object compares equal *this, allocating array object of same size , type.

which means size must match precisely, otherwise run undefined behavior.


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