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
This is not about associativity. This is about order of evaluation of sub-expressions. — T.C.
C++11 (draft N3337) dcl.fct.default: The order of evaluation of function arguments is unspecified. — legends2k

AnT AnT · Accepted Answer · 2015-01-22T01:21:28

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.

Precedence of operators?

2 Answers