This doesn't compile:
bool IsDivisible(int x, int y) {
return x % y == 0;
}
...
EXPECT_CALL(...)
.With(Args<0, 1>(Truly(&IsDivisible));
as Truly() expects its argument to be a function that takes exactly one
argument (in this case, a 2-tuple).
The user needs to write IsDivisible() as
bool IsDivisible(const tr1::tuple<int, int> p) {
return tr1::get<0>(p) % tr1::get<1>(p) == 0;
}
which is tedious and unobvious.
The original definition of IsDivisible should just work in the example.
Original issue reported on code.google.com by w...@google.com on 9 Sep 2010 at 5:21
Original issue reported on code.google.com by
w...@google.com
on 9 Sep 2010 at 5:21