Does an expression with undefined behaviour that is never actually executed make a program erroneous?

Question

In many discussions about undefined behavior (UB), the point of view has been put forward that in the mere presence in a program of any construct that has UB in a program mandates a conforming implementation to do just anything (including nothing at all). My question is whether this should be taken in that sense even in those cases where the UB is associated to the execution of code, while the behaviour (otherwise) specified in the standard stipulates that the code in question should not be executed (and this possibly for specific input to the program; it might not be decidable at compile time).

Phrased more informally, does the smell of UB mandate a conforming implementation to decide that the whole program stinks, and refuse to execute correctly even the parts of the program for which the behaviour is perfectly well defined. An example program would be

#include <iostream>

int main()
{
    int n = 0;
    if (false)
      n=n++;   // Undefined behaviour if it gets executed, which it doesn't
    std::cout << "Hi there.\n";
}

For clarity, I am assuming the program is well-formed (so in particular the UB is not associated to preprocessing). In fact I am willing to restrict to UB associated to "evaluations", which clearly are not compile-time entities. The definitions pertinent to the example given are, I think,(emphasis is mine):

Sequenced before is an asymmetric, transitive, pair-wise relation between evaluations executed by a single thread (1.10), which induces a partial order among those evaluations

The value computations of the operands of an operator are sequenced before the value computation of the result of the operator. If a side effect on a scalar object is unsequenced relative to either ... or a value computation using the value of the same scalar object, the behavior is undefined.

It is implicitly clear that the subjects in the final sentence, "side effect" and "value computation", are instances of "evaluation", since that is what the relation "sequenced before" is defined for.

I posit that in the above program, the standard stipulates that no evaluations occur for which the condition in the final sentence is satisfied (unsequenced relative to each other and of the described kind) and that therfore the program does not have UB; it is not erroneous.

In other words I am convinced that the answer to the question of my title is negative. However I would appreciate the (motivated) opinions of other people on this matter.

Maybe an additional question for those who advocate an affirmative answer, would that mandate that the proverbial reformatting of your hard drive might occur when an erroneous program is compiled?

Some related pointers on this site:

• Observable behavior and undefined behavior -- What happens if I don't call a destructor?
• Comments to this answer https://stackoverflow.com/a/24143792/1436796 (I do no longer stand absolutely with my answer itself)
• C++ What is the earliest undefined behavior can manifest itself?
• Difference between Undefined Behavior and Ill-formed, no diagnostic message required and its two answers, which represent opposite points of view

Answer 1

If a side effect on a scalar object is unsequenced relative to etc

Side effects are changes in the state of the execution environment (1.9/12). A change is a change, not an expression that, if evaluated, would potentially produce a change. If there is no change, there is no side effect. If there is no side effect, then no side effect is unsequenced relative to anything else.

This does not mean that any code which is never executed is UB-free (though I'm pretty sure most of it is). Each occurrence of UB in the standard needs to be examined separately. (The stricken-out text is probably overly cautious; see below).

The standard also says that

A conforming implementation executing a well-formed program shall produce the same observable behavior as one of the possible executions of the corresponding instance of the abstract machine with the same program and the same input. However, if any such execution contains an undefined operation, this International Standard places no requirement on the implementation executing that program with that input (not even with regard to operations preceding the first undefined operation).

(emphasis mine)

This, as far as I can tell, is the only normative reference that says what the phrase "undefined behavior" means: an undefined operation in a program execution. No execution, no UB.

Answer 2

No. Example:

struct T {
    void f() { }
};
int main() {
    T *t = nullptr;
    if (t) {
        t->f(); // UB if t == nullptr but since the code tested against that
    }
}