Does size_t have the same size and alignment as ptrdiff_t?
In short: is it required by the standard?
No. The only requirement is from [support.types.layout]/2 and it is:
The type ptrdiff_t is an implementation-defined signed integer type that can hold the difference of two subscripts in an array object, as described in [expr.add].
There is paragraph 4
[ Note: It is recommended that implementations choose types for ptrdiff_t and size_t whose integer conversion ranks are no greater than that of signed long int unless a larger size is necessary to contain all the possible values. — end note ]
but notes are non-normative and it is only a recommendation, not a requirement.
std::size_t is defined as
The type size_t is an implementation-defined unsigned integer type that is large enough to contain the size in bytes of any object ([expr.sizeof]).
in paragraph 3 and it also has no requirement that they be the same.
It is not required by the standard.
Note that the current crop of Intel processors have 48 bit pointers under the hood.
So personally I don't see it too far-fetched to conceive a 64 bit unsigned for std::size_t and a 49 bit signed type for a std::ptrdiff_t. Although such a scheme would be a headache to implement.
More interestingly once chipsets evolve to have 64 bit pointers (we are some way off that being necessary), presumably std::ptrdiff_t will have to be at least 65 bits! Personally therefore I keep in mind that one day sizeof(std::ptrdiff_t) may be larger than sizeof(std::size_t).