Are nullptr references undefined behaviour in C++? [duplicate]

Question

This question already has answers here:

Assigning a reference by dereferencing a NULL pointer (5 answers)

Closed 8 years ago.

The following code fools around with nullptr pointer and reference:

#include <cstdio>

void printRefAddr(int &ref) {
    printf("printAddr %p\n", &ref);
}

int main() {    
    int *ip = nullptr;
    int &ir = *ip;

    // 1. get address of nullptr reference
    printf("ip=%p &ir=%p\n", ip, &ir);

    // 2. dereference a nullptr pointer and pass it as reference
    printRefAddr(*ip);

    // 3. pass nullptr reference
    printRefAddr(ir);

    return 0;
}

Question: In C++ standards, are commented statements 1..3 valid code or undefined behavior?

Is this same or different with different versions of C++ (older ones would of course use 0 literal instead of nullptr keyword)?

Bonus question: are there known compilers / optimization options, which would actually cause above code to do something unexpected / crash? For example, is there a flag for any compiler, which would generate implicit assertion for nullptr everywhere where reference is initialized, including passing reference argument from *ptr?

---

An example output for the curious, nothing unexpected:

ip=(nil) &ir=(nil)
printAddr (nil)
printAddr (nil)

Answer 1

// 2. dereference a nullptr pointer and pass it as reference

Dereferencing a null pointer is Undefined Behaviour, and so whether you pass it as a reference or by value, the fact is that you've dereferenced it and therefore invoked UB, meaning from that point on all bets are off.

You've already invoked UB here:

int &ir = *ip; //ip is null, you cannot deref it without invoking UB.

Answer 2

Since ir is just a shadow of *ip it will not cause an undefined behavior on its own.

The undefined behavior is using a pointer which points to nullptr_t. I mean using *ip. Therefore

int &ir = *ip;
          ^^^

Causes an UB.