Is the following an undefined behavior? i = func(i)

Question

I know that i=i++; is an undefined behavior, because i is changed twice before the sequence point ;.

But I don't know if the compiler guarantees the case as below is not an undefined behavior:

int func(int &i)
{
    i++;
    return i;
}

int i = 1;
i = func(i);

Answer 1

Firstly, modern C++ has switched from the old (inadequate) concept of "sequence points" to the new concept of "sequencing" (i.e. "sequenced before", "sequenced after"). While i = i++ is still undefined, i = ++i is actually perfectly defined now. Sequencing rules in many lvalue-returning operators were reworked.

Secondly, your version is safe under the old specification as well as under the new one. The modification of i inside the function is safely "isolated" from the assignment to i outside. In the classic specification sequence points at the beginning and at the end of the function safely separated the modifications (and reads) of i from each other. The new sequencing rules preserve the same level of protection as well.

An example that illustrates the protection provided by a function call might look as follows

int inc(int &i) { return i++; }
...
int i = 1;

int r1 = i++ * i++ * i++;          
// Undefined behavior because of multiple unsequenced side effects
// applied to the same variable

int r2 = inc(i) * inc(i) + inc(i);
// No UB, but order of evaluation is unspecified. Since the result 
// depends on the order of evaluation, it is unspecified

int r3 = inc(i) + inc(i) + inc(i); 
// Perfectly defined result. Order of evaluation is still unspecified, 
// but the result does not depend on it