Question

Recently in an interview there was a following objective type question.

int a = 0;
cout << a++ << a;

Answers:

a. 10
b. 01
c. undefined behavior

I answered choice b, i.e. output would be “01”.

But to my surprise later I was told by an interviewer that the correct answer is option c: undefined.

Now, I do know the concept of sequence points in C++. The behavior is undefined for the following statement:

int i = 0;
i += i++ + i++;

but as per my understanding for the statement cout << a++ << a , the ostream.operator<<() would be called twice, first with ostream.operator<<(a++) and later ostream.operator<<(a).

I also checked the result on VS2010 compiler and its output is also ’01’.

Answer 1

You can think of:

cout << a++ << a;

As:

std::operator<<(std::operator<<(std::cout, a++), a);

C++ guarantees that all side effects of previous evaluations will have been performed at sequence points. There are no sequence points in between function arguments evaluation which means that argument a can be evaluated before argument std::operator<<(std::cout, a++) or after. So the result of the above is undefined.

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C++17 update

In C++17 the rules have been updated. In particular:

In a shift operator expression E1<<E2 and E1>>E2, every value computation and side-effect of E1 is sequenced before every value computation and side effect of E2.

Which means that it requires the code to produce result b, which outputs 01.

See P0145R3 Refining Expression Evaluation Order for Idiomatic C++ for more details.