Assignment between union members

Question

Is this code well-defined?

int main()    
{    
    union     
    {    
        int i;    
        float f;    
    } u;    

    u.f = 5.0;    
    u.i = u.f;       // ?????
}    

It accesses two different union members in one expression so I am wondering if it falls foul of the [class.union]/1 provisions about active member of a union.

The C++ Standard seems to underspecify which operations change the active member for builtin types, and what happens if an inactive member is read or written.

Answer 1

The assignment operator (=) and the compound assignment operators all group right-to-left. [...] In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression. [...]

[N4431 §5.18/1]

Under the premise that in your case the value computation of the left hand side (involving merely a member access, not a read) does not cause undefined behaviour, I'd apply the above as follows:

1. The value computation of the right side reads u.f, which is the active member of the union. So everything is good.
2. The assignment is executed. This involves writing the obtained result to u.i, which now changes the active member of the union.