(I know what the scope resolution operator does, and how and when to use it.)
Why does C++ have the ::
operator, instead of using the .
operator for this purpose? Java doesn't have a separate operator, and works fine. Is there some difference between C++ and Java that means C++ requires a separate operator in order to be parsable?
My only guess is that ::
is needed for precedence reasons, but I can't think why it needs to have higher precedence than, say, .
. The only situation I can think it would is so that something like
a.b::c;
would be parsed as
a.(b::c);
, but I can't think of any situation in which syntax like this would be legal anyway.
Maybe it's just a case of "they do different things, so they might as well look different". But that doesn't explain why ::
has higher precedence than .
.
Because someone in the C++ standards committee thought that it was a good idea to allow this code to work:
struct foo
{
int blah;
};
struct thingy
{
int data;
};
struct bar : public foo
{
thingy foo;
};
int main()
{
bar test;
test.foo.data = 5;
test.foo::blah = 10;
return 0;
}
Basically, it allows a member variable and a derived class type to have the same name. I have no idea what someone was smoking when they thought that this was important. But there it is.
When the compiler sees .
, it knows that the thing to the left must be an object. When it sees ::
, it must be a typename or namespace (or nothing, indicating the global namespace). That's how it resolves this ambiguity.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With