pointer cast

[Carlos F.]

Hi everyboy, somebody could tell me what meen the following cast:

a = (struct foo 8) b
c = int (func) (int)

Thanks

Carlos

[Christopher Benson-Manica]

Re: pointer cast

Carlos F. <prohto@yahoo.c om> spoke thus:
[color=blue]
> a = (struct foo 8) b[/color]

If you meant (struct foo *) b, it's pretty self explanatory - cast b
to a foo pointer and assign it to a. Whether it results in
well-defined behavior or not depends on what a and b are declared as
and what's happened prior to the line in question.
[color=blue]
> c = int (func) (int)[/color]

Depends on what func is, and it's probably illegal anyway unless func
is an obfuscated macro. Post more code.

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[Karthik]

Re: pointer cast

Carlos F. wrote:
[color=blue]
> Hi everyboy, somebody could tell me what meen the following cast:
>
> a = (struct foo 8) b[/color]
Did u mean -
a = (struct foo *) b

Then you are casting pointer to type b to pointer to type a.
[color=blue]
> c = int (func) (int)[/color]
Not sure with this syntax .. Do you want to check that ?


[color=blue]
>
> Thanks
>
> Carlos[/color]


--
Karthik.
Humans please 'removeme_' for my real email.

[CBFalconer]

Re: pointer cast

"Carlos F." wrote:[color=blue]
>
> Hi everyboy, somebody could tell me what meen the following cast:
>
> a = (struct foo 8) b[/color]

This is a syntax error.
[color=blue]
> c = int (func) (int)[/color]

If func exists both as a functional macro, and as a callable
function, this forces the actual function to be called. The macro
can only be called by "func(" which this is not.

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