screwed here!!!!!

ayan_2587 · 11-11-2009, 01:37 AM

Code:

int fun(int(*)());

int main()
{
   fun(main);
   cout<<"in main"<<endl;
   return 0;
}

fun(int (*p)())
{
  cout<<"in fun"<<endl;
  return 0;
}

the output is....

in fun
in main

how and in which manner is "main" being passed in the fun function argument in main function.
does this way of passing automatically mean that a function pointer that returns an integer is being passed into the fun function argument????

how does this program work..
kindly help..
Thnx

jdragyn · 11-11-2009, 01:52 AM

What are you expecting? You would get the same output if you wrote

Code:

int fun(int);

int main()
{
   fun(42);
   cout<<"in main"<<endl;
   return 0;
}

fun(int x)
{
  cout<<"in fun"<<endl;
  return 0;
}

RockyMarrone · 11-11-2009, 02:14 AM

This is as simple as calling a normal function.. with the code above it will call fun and pass a function pointer with which you are not doing anything

. Than it will call main cout

leeor_net · 11-11-2009, 02:19 AM

Exactly what are you trying to do with your code? I've never seen a recursive call to an entry point function (main) let a lone a question that doesn't really ask anything.

If you're expecting to see "in main" print before "in fun" than you've misunderstood function pointers and function calls altogether. You'd need to call main via your function pointer and, if you tried to do that with this program you'd end up in an infinte loop if it even worked at all.

In this case, as you've not called main() via your function pointer your function fun() simply prints out a statement and then returns where main continues merily along.

RockyMarrone · 11-11-2009, 02:34 AM

Quote:

Originally Posted by leeor_net

Exactly what are you trying to do with your code? I've never seen a recursive call to an entry point function (main) let a lone a question that doesn't really ask anything.

If you're expecting to see "in main" print before "in fun" than you've misunderstood function pointers and function calls altogether. You'd need to call main via your function pointer and, if you tried to do that with this program you'd end up in an infinte loop if it even worked at all.

In this case, as you've not called main() via your function pointer your function fun() simply prints out a statement and then returns where main continues merily along.

Hey dude what are you talkin abt i don't think so that he is recursively calling the main function he is just passing the pointer to the main function just not doing any thing with that pointer

rags_to_riches · 11-11-2009, 08:43 AM

fun is defined as taking as its argument a pointer to a function which takes no arguments and returns an integer...that's what this

Code:

int (*p)()

means.

the name of a function, without any brackets(), is a pointer to the function with that name. Hence, main is a pointer to the function

Code:

int main()

grumpy · 11-11-2009, 10:35 AM

Quote:

Originally Posted by rags_to_riches

the name of a function, without any brackets(), is a pointer to the function with that name.

Not true. The name of a function is not a pointer. It can be converted by the compiler into a pointer to a function in some contexts.

Quote:

Originally Posted by rags_to_riches

Hence, main is a pointer to the function

Code:

int main()

main() is a very special case in C++ - a pointer to main() cannot be created. This is related to the fact that main() cannot be called recursively.

rags_to_riches · 11-11-2009, 11:31 AM

Sorry, I'm not really a language lawyer.

11-11-2009, 01:37 AM	#1
ayan_2587 Registered User Join Date: Oct 2009 Posts: 11	screwed here!!!!! Code: int fun(int()()); int main() { fun(main); cout<<"in main"<<endl; return 0; } fun(int (p)()) { cout<<"in fun"<<endl; return 0; } the output is.... in fun in main how and in which manner is "main" being passed in the fun function argument in main function. does this way of passing automatically mean that a function pointer that returns an integer is being passed into the fun function argument???? how does this program work.. kindly help.. Thnx
ayan_2587 is offline

11-11-2009, 01:52 AM	#2
jdragyn Registered User Join Date: Sep 2009 Posts: 45	What are you expecting? You would get the same output if you wrote Code: int fun(int); int main() { fun(42); cout<<"in main"<<endl; return 0; } fun(int x) { cout<<"in fun"<<endl; return 0; }
jdragyn is offline

11-11-2009, 02:14 AM	#3
RockyMarrone Registered User Join Date: Oct 2009 Location: While(1) Posts: 368	This is as simple as calling a normal function.. with the code above it will call fun and pass a function pointer with which you are not doing anything . Than it will call main cout
RockyMarrone is offline

11-11-2009, 02:19 AM	#4
leeor_net Professional Chef Join Date: Apr 2004 Location: Charles Town, WV Posts: 144	Exactly what are you trying to do with your code? I've never seen a recursive call to an entry point function (main) let a lone a question that doesn't really ask anything. If you're expecting to see "in main" print before "in fun" than you've misunderstood function pointers and function calls altogether. You'd need to call main via your function pointer and, if you tried to do that with this program you'd end up in an infinte loop if it even worked at all. In this case, as you've not called main() via your function pointer your function fun() simply prints out a statement and then returns where main continues merily along. __________________ - Leeor Last edited by leeor_net; 11-11-2009 at 02:24 AM.
leeor_net is offline

11-11-2009, 08:43 AM	#6
rags_to_riches and the Hat of Ass Join Date: Dec 2007 Posts: 813	fun is defined as taking as its argument a pointer to a function which takes no arguments and returns an integer...that's what this Code: int (*p)() means. the name of a function, without any brackets(), is a pointer to the function with that name. Hence, main is a pointer to the function Code: int main()
rags_to_riches is offline

11-11-2009, 11:31 AM	#8
rags_to_riches and the Hat of Ass Join Date: Dec 2007 Posts: 813	Sorry, I'm not really a language lawyer.
rags_to_riches is offline

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