Pointer arithmetics with two different buffers

Consider the following code:

int* p1 = new int[100];
int* p2 = new int[100];
const ptrdiff_t ptrDiff = p1 - p2;

int* p1_42 = &(p1[42]);
int* p2_42 = p1_42 + ptrDiff;

Now, does the Standard guarantee that p2_42 points to p2[42]? If not, is it always true on Windows, Linux or webassembly heap?

To add the standard quote:

expr.add#5

When two pointer expressions P and Q are subtracted, the type of the result is an implementation-defined signed integral type; this type shall be the same type that is defined as std::ptrdiff_­t in the <cstddef> header ([support.types]).

• (5.1) If P and Q both evaluate to null pointer values, the result is 0.

• (5.2) Otherwise, if P and Q point to, respectively, elements x[i] and x[j] of the same array object x, the expression P - Q has the value i−j.

• (5.3) Otherwise, the behavior is undefined. [ Note: If the value i−j is not in the range of representable values of type std::ptrdiff_­t, the behavior is undefined. — end note  ]

(5.1) does not apply as the pointers are not nullptrs. (5.2) does not apply because the pointers are not into the same array. So, we are left with (5.3) - UB.

const ptrdiff_t ptrDiff = p1 - p2;

This is undefined behavior. Subtraction between two pointers is well defined only if they point to elements in the same array. ([expr.add] ¶5.3).

When two pointer expressions P and Q are subtracted, the type of the result is an implementation-defined signed integral type; this type shall be the same type that is defined as std::ptrdiff_­t in the <cstddef> header ([support.types]).

• If P and Q both evaluate to null pointer values, the result is 0.
• Otherwise, if P and Q point to, respectively, elements x[i] and x[j] of the same array object x, the expression P - Q has the value i−j.
• Otherwise, the behavior is undefined

And even if there was some hypothetical way to obtain this value in a legal way, even that summation is illegal, as even a pointer+integer summation is restricted to stay inside the boundaries of the array ([expr.add] ¶4.2)

When an expression J that has integral type is added to or subtracted from an expression P of pointer type, the result has the type of P.

• If P evaluates to a null pointer value and J evaluates to 0, the result is a null pointer value.
• Otherwise, if P points to element x[i] of an array object x with n elements,⁸¹ the expressions P + J and J + P (where J has the value j) point to the (possibly-hypothetical) element x[i+j] if 0≤i+j≤n and the expression P - J points to the (possibly-hypothetical) element x[i−j] if 0≤i−j≤n.
• Otherwise, the behavior is undefined.