c++ – About the Order of Expression Evaluation

Question:

I would like to understand what rules determine the order of evaluation of the values ​​of expressions in the general case.

Let's say there is a code like this

int readValue()
{
  int v;
  cin >> v;
  return v;
}

int main()
{
  cout << readValue() << ' ' << readValue() << '\n';
  return 0;
}

As you know , the bitwise shift operator is calculated from left to right – but when you enter 1 2 , the output is 2 1 (Microsoft's compiler), what is the reason for this?

Does this code generate undefined behaviour or is it just the output order is undefined ?

If you remove the reading code – everything is pretty predictable

int readValue(int v)
{
  return v;
}

int main()
{
  cout << readValue(1) << ' ' << readValue(2) << '\n';
  return 0;
}

output – 1 2

what's the matter ?

Answer:

The statement that these bitwise shift operators are evaluated from left to right is true, but it only tells you the order in which successive << operators are applied to their immediate operands. In this case, the direct operands of the << operators are some temporary intermediate values ​​of the int type

cout << __tmp_int1 << ' ' << __tmp_int2 << '\n';

Here about this (and only about this) expression in this case, we can say that it is calculated from left to right. That is, the value of __tmp_int1 will be printed earlier, and the value of __tmp_int2 – later.

And as for the order and moment of preparation of these intermediate values ​​- it is not limited to the above statement. That is, the compiler can do and

int __tmp_int1 = ReadValue();
int __tmp_int2 = ReadValue();

and vice versa

int __tmp_int2 = ReadValue();
int __tmp_int1 = ReadValue();

or generally intertwine the preparation __tmp_int1 and __tmp_int2 with << calls in some unpredictable way. The main thing is that the value is ready for the moment when it is needed. Nothing more is specified.