Re: Is there a way to remove the requirement for parenthesis?

[Charles Hixson]

Sergey Gromov wrote:
Wed, 21 Jan 2009 09:24:01 -0800, Charles Hixson wrote:
In this test I'm trying to emulate how I want a typedef to act, but I 
run into a problem:

import   std.stdio;

struct   BlockNum
{  uint   value;

    uint   opCast()   {   return   value;   }
    void   opAssign (uint val)   {   value = val;   }
    uint   opCall()   {   return   value;   }
}

void   main()
{  BlockNum   test;
    test   =   42;
    uint   tst2   =   test();  // <<== if I don't have the parenthesis I
                               //    get a compiler error (cast
                               //    required).
           //  kfile.d(15): Error: cannot implicitly convert expression
           //          (test) of type BlockNum to uint

    writef ("tst2 = %d\n", tst2);
}

It seemed to me as if the parens shouldn't be required here, but I seem 
mistaken.  Which leads to ugly code.  Is there a way around this?

test is an expression of type BlockNum.  opCall() is called when you use
parentheses syntax on it.  opCast() is called when you use cast() syntax
for it.  Otherwise it stays BlockNum and therefore is not convertible to
uint.
I think that means "No, there isn't a way around it."
OK.  I'll just ...
well, I'm not totally sure.  Either give up type safety or ... something 
else.