If we execute the code fragment in an otherwise complete,correct program:
n = 1;
cout

Question

If we execute the code fragment in an otherwise complete,correct program:
n = 1;
cout << n++ << " " << n++ << " " << n++ << endl;
the output is guaranteed to be 1 2 3.

Quizplus · Accepted Answer

The order of evaluation of function arguments is unspecified in C++, and the increment operator changes the value of the operand. Therefore, the output of the code is not guaranteed to be 1 2 3. It may be any sequence of three consecutive integers depending on the order of evaluation.

If We Execute the Code Fragment in an Otherwise Complete,correct