Nice program, division by zero exception :)
>Hi all ,
>
> This was my solution.
>
> #include <stdio.h>
>#include <stdlib.h>
>#include <setjmp.h>
>static jmp_buf buf;
>int main( int argc , char **argv )
>{
> int i=0;
>
>
> try {
>setjmp(buf);
> printf("%d\n",i++);
> longjmp(buf,i/(i%300001));
> }
> catch(...) {}
>
>}
>
>regards
>Praseed Pai
>P.S :- Template MetaProgramming solution seems to be good. It is a kind of
>loop
>unrolling and it is also >called Compile time recursion.
--- On Thu, 3/31/11, Faisal Fiza <[email protected]> wrote:
>From: Faisal Fiza <[email protected]>
>Subject: Re: [ILUG-Cochin.org] 300000 problem
>To: "This List discusses GNU/Linux & GNU, GPL Software "
><[email protected]>
>Date: Thursday, March 31, 2011, 11:44 AM
>
>
>Hehe :)
>
>its good but the 'seq' program may contain loops and conditions.
>
>finally i got the exact link which expected.
>
>http://stackoverflow.com/questions/4568645/printing-1-to-1000-without-loop-or-conditionals
>
>
>
>
>
>
>
>
>>Althaf's example is good, based on your conditions. You can call an external
>>program
>>to do this in GNU. The following will print it.
>>
>>
>>int main()
>>{
>> system("seq 30000");
>> }
>>
>> Regards,
>> Jos Collin
>
>>On Thu, Mar 31, 2011 at 07:59:01PM +0530, Faisal Fiza wrote:
>> >
>> >
>> >
>> >
>> >
>> >
>> > Recursion still has conditionals to check the exit condition, however goto
>> > is
>>
>>
>> >not exactly a loop, moreover in > the question it doesnt mention program
>>should
>>
>> >terminate once 0 to 3.... is printed, ( word play :-D), either >
>> >recursion
>>or
>>
>> >goto will do, recursion will end up in stack over flow, goto will end up in
>> >infinite loop. ;-).
>>>
>>>
>>> Conditions:
>>> *Program should terminate after printed from 0 to 3lak.
>>> *No conditions and loops
>>> *If u use assembly , don't use conditional jumps.
>
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>
>
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