saroj@bear.com wrote:
>
> Hi,
>
> I am trying to do some pretty-printing by parsing C++ code. I only
> know if a name is a type or not. So int and X (where X is a class name)
> are equivalent.
> Given a code fragment:
>
>   X :: name ...
>
> how can I know if :: is part of X or if it is part of name?
> e.g. struct X { typedef int Y;};
>
> X::Y i;  // Y is a typedef in class X (X::Y printed together)
> X ::Y::f();  // f is a function in namespace Y returning a X.
>              // (::Y::f is printed together)

Good question. The standard is not clear on the issue of how
to parse a declaration when the declarator starts with a
global scope modifier, as in your example. It is currently
a subject of discussion in the C++ Committee.

Some compilers apply the "maximum munch" rule, and parse the
second line as
 X::Y::f(); // call function X::Y::f

If you are writing code, you should use parentheses around the
declarator to disambiguate:
 X ( ::Y::f )(); // declare function ::Y::f
 X ( ::Y::f() ); // another way to parenthesize
 ( X::Y::f )();  // call function X::Y::f
 ( X::Y::f() );  // another way to parenthesize

In these examples I'm using spaces to help show how the
tokens go together; the spaces have no syntactic meaning.

Note that you cannot put parens around the type:
 (X) ::Y::f(); // syntax error

--
Steve Clamage, stephen.clamage@sun.com
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Hi,

I am trying to do some pretty-printing by parsing C++ code. I only
know if a name is a type or not. So int and X (where X is a class name)
are equivalent.
Given a code fragment:

  X :: name ...

how can I know if :: is part of X or if it is part of name?
e.g. struct X { typedef int Y;};

X::Y i;  // Y is a typedef in class X (X::Y printed together)
X ::Y::f();  // f is a function in namespace Y returning a X.
             // (::Y::f is printed together)

Thank you,
Saroj Mahapatra


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