Pointer arithmetics with two different buffers

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Category:Languages

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.

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