RE: Function call

RE: Function call

[Honnavalli_Sreevathsa]

The statement 

funckey;

does not have any effect and the compiler does not generate any extra
assembly code for it. You are just mentioning a valid symbol name in your
program without assigning it to any other variable, which the compiler deems
as 'not wrong'. You can compile your program with -Wall option to see a
warning message saying this.

Sreevathsa



> -----Original Message-----
> From: linux-c-programming-owner@vger.kernel.org 
> [mailto:linux-c-programming-owner@vger.kernel.org] 
> Sent: Wednesday, March 01, 2006 10:54 AM
> To: Linux C Programming List
> Subject: Function call 
> 
> 
> Hello,
> 
> I have a doubt regarding the following code. There is a void function
> funckey().
> 
> void funckey()
> {
> ;
> }
> 
> void main()
> {
> funckey;
> funckey();
> }
> 
> During execution, the second statement calls the function funckey().
> 
> What does the compiler do in the first statement funckey?
> 
> Thanks in advance, 
> Subbulakshmi Sadagopal 
> 
> -
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Re: Function call

[Shriramana Sharma]

Wednesday, 01 March 2006 16:17 samaye, Honnavalli_Sreevathsa@emc.com alekhiit:

> The statement funckey; 'not wrong'. 
> You can compile your program with -Wall option to see a warning message
> saying this. 

fk.c:8: warning: statement with no effect

-- 

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$ date [] CCE +2006-03-01 W09-3 UTC+0530

Re: Function call

[wwp]

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Hello Shriramana,


On Wed, 1 Mar 2006 16:24:04 +0530 Shriramana Sharma <samjnaa@gmail.com> wrote:

> Wednesday, 01 March 2006 16:17 samaye, Honnavalli_Sreevathsa@emc.com
> alekhiit:
> 
> > The statement funckey; 'not wrong'. 
> > You can compile your program with -Wall option to see a warning message
> > saying this. 
> 
> fk.c:8: warning: statement with no effect

I'm not sure w/ gcc, but a "statement with no effect" might not mean that the
statement isn't compiled and generates code.. that will have no effect ;-).


Regards,

-- 
wwp

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Re: Function call

[Glynn Clements]

wwp wrote:

> > > The statement funckey; 'not wrong'. 
> > > You can compile your program with -Wall option to see a warning message
> > > saying this. 
> > 
> > fk.c:8: warning: statement with no effect
> 
> I'm not sure w/ gcc, but a "statement with no effect" might not mean that the
> statement isn't compiled and generates code.. that will have no effect ;-).

Without optimisation, it might generate some code (e.g. putting the
value funckey into a register). With optimisation, wouldn't expect any
code to be generated.

-- 
Glynn Clements <glynn@gclements.plus.com>

Function call

[Subbulakshmi Sadagopal (RBIN/EDS2) *]

Hello,

I have a doubt regarding the following code. There is a void function
funckey().

void funckey()
{
;
}

void main()
{
funckey;
funckey();
}

During execution, the second statement calls the function funckey().

What does the compiler do in the first statement funckey?

Thanks in advance, 
Subbulakshmi Sadagopal 

Re: Function call

[wwp]

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Hello Subbulakshmi,


On Wed, 1 Mar 2006 10:54:03 +0530 "Subbulakshmi Sadagopal (RBIN/EDS2) *" <Subbulakshmi.S@in.bosch.com> wrote:

> 
> Hello,
> 
> I have a doubt regarding the following code. There is a void function
> funckey().
> 
> void funckey()
> {
> ;
> }
> 
> void main()
> {
> funckey;
> funckey();
> }
> 
> During execution, the second statement calls the function funckey().
> 
> What does the compiler do in the first statement funckey?

It evaluates the expression "funckey", which is just a pointer to a function
(as char* foo; is a pointer to char). Function names can be used in some few
situations (passed as parameter to some functions for signal/callback
hooking for instance).


Regards,

-- 
wwp

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