Precedence of operators?

Question

Look at this simple class:

class A {
    int *val;
public:
    A() { val = new int; *val = 0; }
    int get() { return ++(*val); }
};

Why when I run this code it prints 21:

int main() {
    A a, b = a;
    cout << a.get() << b.get();
    return 0;
}

But if I run it like this it prints 12 which is what I was expecting:

int main() {
    A a, b = a;
    cout << a.get();
    cout << b.get();
    return 0;
}

What am I missing here? Operator precedence? FYI, this is a C++ test problem, not a production code.

EDIT: Does it means that when I have cout << (Expr1) << (Expr2) then Expr1 and Expr2 are evaluated before the output of Expr1 is printed?

This is about associativity, not precedence. There is only one operator, so the question of precedence cannot arise. — user207421, Jan 22 '15 at 01:07
The first one could print `12` also depending on the compiler you use. — Praetorian, Jan 22 '15 at 01:08
This is not about associativity. This is about order of evaluation of sub-expressions. — T.C., Jan 22 '15 at 01:08
@T.C. Which is determined by associativity, if it's defined at all. — user207421, Jan 22 '15 at 01:09
http://stackoverflow.com/questions/5473107/operator-precedence-vs-order-of-evaluation — T.C., Jan 22 '15 at 01:14
C++11 (draft N3337) _dcl.fct.default_: `The order of evaluation of function arguments is unspecified.` — legends2k, Jan 22 '15 at 01:14

AnT stands with Russia · Accepted Answer · 2015-01-22T01:26:40.323

Operator precedence does not dictate the order of evaluation of intermediate results. I.e. it does not dictate the order of evaluation of sub-expressions in the entire expression.

Expression statement

cout << a.get() << b.get();

can be executed by the compiler as

int tmp_a = a.get();
int tmp_b = b.get();
cout << tmp_a;
cout << tmp_b;

or as

int tmp_b = b.get();
int tmp_a = a.get();
cout << tmp_a;
cout << tmp_b;

In this case operator precedence and associativity for operator << guarantees that tmp_a is sent to output before tmp_b. But it does not guarantee that tmp_a is evaluated before tmp_b.

Drew Dormann · Answer 2 · 2015-01-22T01:21:50.297

2

This line:

cout << a.get() << b.get();

does not specify whether a.get() or b.get() is evaluated first.

Since a and b share the same val pointer, the order of those operations affect the output.

(Edit)

Does it means that when I have cout << (Expr1) << (Expr2) then Expr1 and Expr2 are evaluated before the output of Expr1 is printed?

That is also not specified.

edited Jan 22 '15 at 01:21

answered Jan 22 '15 at 01:07

Drew Dormann

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Done! However I'm still not clear why `Expr2` is being evaluated before `Expr1`. I have tried GCC and MSVC – Darien Pardinas Jan 22 '15 at 01:21
@DarienPardinas it's because they want to, and they're allowed to. You, as a programmer, are not to assume which will be evaluated first. And if you want that control, you can put them on separate lines. Good luck! – Drew Dormann Jan 22 '15 at 01:23
@DarienPardinas Don't confused *observed* behavior with *specified* behavior or *defined* behavior. They're sometimes different things. When in doubt peel back the standard text and see where it takes you. – WhozCraig Jan 22 '15 at 01:26
Thanks! Anyway, that was one of the practice questions of the C++ Certified Associate Programmer and the right answer according to them is `21` even when that output should be undefined. – Darien Pardinas Jan 22 '15 at 01:27
2

@DarienPardinas Wow! Someone should have a chat with them, no? Can you share a link? – Drew Dormann Jan 22 '15 at 01:28
http://education.cppinstitute.org/ Chapter 6 Question 7. You need to register (it's free) to see it with your own eyes. – Darien Pardinas Jan 22 '15 at 01:33
@DarienPardinas their stock photos use differing gaussian blurs. That's the first red flag. – Drew Dormann Jan 22 '15 at 01:36
@DrewDormann What yo you mean with that? That is not worth looking at this certification? – Darien Pardinas Jan 22 '15 at 04:58

Precedence of operators?

2 Answers2