Pointer arithmetic with variables not in an array

Question

I have the following question; If a is an int array with 10 Element I can define pointers

int*b=&a[3]; 
int*c=&[2];

I can then do arithmetic operations with these pointers like int d=a-c; which will return the number of int values in the array between b and c. So my question is if I am also allowed to do such pointer arithmetic operations for any variables which may not be in an array. For example:

int a=10; 
int b=20; 
int*c=&a; 
int* d=&b;

and then do int e=d-c; or int*e=c+1;

The reason I ask is that I have received conflicting information about whether this leads to undefined behaviour,

Welcome to stackoverflow.com. Please take some time to read the help pages, especially the sections named "What topics can I ask about here?" and "What types of questions should I avoid asking?". Also please take the tour and read about how to ask good questions. Lastly please learn how to create a Minimal, Complete, and Verifiable Example. — Some programmer dude
– Some programmer dude, Commented Jan 18, 2018 at 13:40
Well, here's the conflict breaker. You may not. Doing so would be UB. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Jan 18, 2018 at 13:40
intb=&a[3] and intc=&a[2] are not not syntatically correct. — Arnav Borborah
– Arnav Borborah, Commented Jan 18, 2018 at 13:41
Ok thanks. Would a comparison if(p1<p2) work for such pointers? — fibo11235
– fibo11235, Commented Jan 18, 2018 at 13:49
Would a comparison if(p1<p2) work for such pointers? it would work on most implementations but it also would be UB. — user2672107
– user2672107, Commented Jan 18, 2018 at 13:53

eerorika · Accepted Answer · 2018-01-18 14:09:42Z

[expr.add] standard draft:

When an expression that has integral type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the expression P points to element x[i] of an array object x with n elements,⁸⁶ 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; otherwise, the behavior is undefined. Likewise, the expression P - J points to the (possibly-hypothetical) element x[i − j] if 0 ≤ i − j ≤ n; otherwise, the behavior is undefined.

When two pointers to elements of the same array object 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 header (21.2). If the expressions 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. [ 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 ]

86) An object that is not an array element is considered to belong to a single-element array for this purpose; see 8.3.1. A pointer past the last element of an array x of n elements is considered to be equivalent to a pointer to a hypothetical element x[n] for this purpose; see 6.9.2.

c+1 is well defined, because it would be pointing one past the "single-element array" that the variable is treated as for the purpose of the quoted rule, and therefore satisfies 0 ≤ 0 + 1 ≤ 1. But it would not be well defined to indirect that pointer, since it past the end of that "array".

d-c has undefined behaviour.

because it would be pointing one past the "single-element array" I don't think you can treat any variable as a single-element array. It is tempting but there doesn't seem to be any permit to do so in the standard.
@n.m. the standard rules can treat a variable as a single-element array and at least this one does. The programmer cannot (except to the extent that the standard allows).

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