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Hi there

I have the following struct:

struct foo {
union {
unsigned __int64 ui64;
char[8] c;
};
};

what i now want to make possible is...

foo a;
foo b;
int c;
foo d;
a = "abcdefg";
b = 12;
d = (foo) c;

so i think i need to globally overload the cast operator, isn't it?

i know how to overload the cast-operator to convert from my struct to
$type, but how do i do it the other way 'round?

thanx for your help,

-.rhavin

".rhavin grobert" <> wrote in news:1169841744.091698.6020
@s48g2000cws.googlegroups.com:

> Hi there
>
> I have the following struct:
>
> struct foo {
> union {
> unsigned __int64 ui64;
> char[8] c;
> };
> };
>
> what i now want to make possible is...
>
> foo a;
> foo b;
> int c;
> foo d;
> a = "abcdefg";
> b = 12;
> d = (foo) c;
>
> so i think i need to globally overload the cast operator, isn't it?

Why not assignment operators?

foo & foo:perator=(__int64 newvalue)
{
ui64 = newvalue;
}

foo & foo:perator=(const char * cp)
{
strncpy(c, cp, sizeof(c));
}

> i know how to overload the cast-operator to convert from my struct to
> $type, but how do i do it the other way 'round?

That would require you to overload operators on int.... can't be done.

..rhavin grobert wrote:
> [..]
> i know how to overload the cast-operator to convert from my struct to
> $type, but how do i do it the other way 'round?

I think you're talking about a parameterised constructor here.

V
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"Andre Kostur" <> wrote in message
news:Xns98C47E9C29755nntpspamkosutrnet@209.135.99. 21...
> ".rhavin grobert" <> wrote in news:1169841744.091698.6020
>
> Why not assignment operators?

Because that won't solve the '(foo)c' piece of code . Constructors are the
way to go (best in combination with assignment operators, but the latter
won't be needed to make it all work).

- Sylvester