function pointers

[Srinivas]

Hi all,
Can anyone describe how the following program works:

#include<stdio. h>

int sum(int ,int);

int main()
{
int (*ptr)(int ,int);
int result;

ptr=sum;

result=test(&** **********&**** ***&********&** *****&*****&sum );
printf("%d\n",r esult);

return 0;
}

int test(int (*ptr)(int ,int ))
{

return ptr(2,3);
}

int sum(int a,int b)
{
return a+b;
}

srinivas

[Michael Mair]

Re: function pointers

Srinivas wrote:[color=blue]
> Hi all,
> Can anyone describe how the following program works:[/color]

Yes. In order to give you the opportunity to formulate
your exact question, I will give a slightly different
version of your code and comment on a couple of things
which are there.
[color=blue]
> #include<stdio. h>[/color]

#include <stdio.h>

White space and its consistent use makes reading code easier.
[color=blue]
> int sum(int ,int);[/color]

It does not hurt to give the prototype in the same form
as the definition:
int sum (int a, int b);

This also gives the compiler opportunity to warn you about
a change in parameter names -- if the parameter name
changes, the semantics sometimes change, too.

You forgot a prototype for test().

[color=blue]
> int main()[/color]

int main (void)
for completeness[color=blue]
> {
> int (*ptr)(int ,int);
> int result;
>
> ptr=sum;[/color]

ptr is not really used, i.e. unnecessary.

[color=blue]
> result=test(&** **********&**** ***&********&** *****&*****&sum );[/color]

Now, what's that?
result = test(sum);
or
result = test(ptr);
do just fine
[color=blue]
> printf("%d\n",r esult);
>
> return 0;
> }
>
> int test(int (*ptr)(int ,int ))
> {
>
> return ptr(2,3);[/color]

This is fine; in order to emphasize that you are working
with a function pointer, you can also use
return (*ptr)(2, 3);
[color=blue]
> }
>
> int sum(int a,int b)
> {
> return a+b;
> }[/color]

Do you want to know something about function pointers?

Cheers
Michael

#include <stdio.h>

int sum (int a, int b);
int test (int (*ptr)(int, int));


int main (void)
{
int result;

result = test(sum);
printf("%d\n", result);

return 0;
}


int test (int (*ptr)(int, int))
{
return (*ptr)(2, 3);
}


int sum (int a, int b)
{
return a + b;
}

--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[junky_fellow@yahoo.co.in]

Re: function pointers


Michael Mair wrote:[color=blue]
> Srinivas wrote:[color=green]
> > Hi all,
> > Can anyone describe how the following program works:[/color]
>
> Yes. In order to give you the opportunity to formulate
> your exact question, I will give a slightly different
> version of your code and comment on a couple of things
> which are there.
>[color=green]
> > #include<stdio. h>[/color]
>
> #include <stdio.h>
>
> White space and its consistent use makes reading code easier.
>[color=green]
> > int sum(int ,int);[/color]
>
> It does not hurt to give the prototype in the same form
> as the definition:
> int sum (int a, int b);
>
> This also gives the compiler opportunity to warn you about
> a change in parameter names -- if the parameter name
> changes, the semantics sometimes change, too.
>
> You forgot a prototype for test().
>
>[color=green]
> > int main()[/color]
>
> int main (void)
> for completeness[color=green]
> > {
> > int (*ptr)(int ,int);
> > int result;
> >
> > ptr=sum;[/color]
>
> ptr is not really used, i.e. unnecessary.
>
>[color=green]
> > result=test(&** **********&**** ***&********&** *****&*****&sum );[/color]
>[/color]
<snip>

I unable to understand how the above statement is parsed ?
Can you please explain this ?