Re: order of operations question

"Andrei Alexandrescu (See Website For Email)" <>
Fri, 14 Sep 2007 07:05:32 CST
<> wrote:

Say you have this line of code:

m << f() << g();

We know the results of f() will be serialized before the results of
g(). But f is not guaranteed to be executed before g, right?

This straightforward question has provoked some debate at work. One
guy states this is equivalent to this example from section 6.2.2 of
Stroustrup's "The C++ Programming Language" (2nd ed.):

int x = f(2) + g(3); // undefined whether f() or g() is called first

Another contends that since you can write the example as:


that f must be executed before g.


It's easy to model what could happen if each expression is assigned to
an explicit temporary.

m << f() << g();


t1 = f();
t2 = g();
t3 = m.operator<<(t1);

There is a dependency that guarantees the first << call happens before
the second. But there's no dependency between t1 and t2, and the C++
compiler can compute them in any order.


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