Yet another fun C-puzzle
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This question was asked in the Trivandrum Microsoft UG forums :- Using C, print 1-100 without using any loop or conditional statements? An answer was later posted in C++ (and the poster said that it was not possible without C++). Here's his C++ solution :-
static int i=0;
class ad
{
ad()
{
i++;
cout << i;
}
}void main()
{
ad objad[100];
}I spent a part of my Sunday evening wondering if it was possible in some alternative manner and found a solution. It shouldn't take you more than 15-20 minutes to think of it (assuming you haven't done this before). Let's see who gets there first (remember - no C++, no classes). Nish :-)
Dunno whether this is acceptable or not ?
static int number = -1;
void PrintNumber()
{
(++number<101)?printf("%d\n",number):Exit(0);PrintNumber();
}int main(int argc, char* argv[])
{
PrintNumber();return 0;}
Imtiaz
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This question was asked in the Trivandrum Microsoft UG forums :- Using C, print 1-100 without using any loop or conditional statements? An answer was later posted in C++ (and the poster said that it was not possible without C++). Here's his C++ solution :-
static int i=0;
class ad
{
ad()
{
i++;
cout << i;
}
}void main()
{
ad objad[100];
}I spent a part of my Sunday evening wondering if it was possible in some alternative manner and found a solution. It shouldn't take you more than 15-20 minutes to think of it (assuming you haven't done this before). Let's see who gets there first (remember - no C++, no classes). Nish :-)
Hmm...
static int i=1;
typedef void (*fn) ();
fn pfn[2];void print_i(void)
{
printf("%d\n",i);
// Not a conditional! :)
// i++ / 100 will evaluate to 0 until i has reached 100
(pfn[i++ / 100])();
}void go_bye_bye()
{
exit(0);
}void main(void)
{
pfn[0] = &print_i;
pfn[1] = &go_bye_bye;(pfn\[0\])();}
-- Russell Morris "So, broccoli, mother says you're good for me... but I'm afraid I'm no good for you!" - Stewy
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Dunno whether this is acceptable or not ?
static int number = -1;
void PrintNumber()
{
(++number<101)?printf("%d\n",number):Exit(0);PrintNumber();
}int main(int argc, char* argv[])
{
PrintNumber();return 0;}
Imtiaz
Isn't the ? : block same as an if-else? Nish
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Hmm...
static int i=1;
typedef void (*fn) ();
fn pfn[2];void print_i(void)
{
printf("%d\n",i);
// Not a conditional! :)
// i++ / 100 will evaluate to 0 until i has reached 100
(pfn[i++ / 100])();
}void go_bye_bye()
{
exit(0);
}void main(void)
{
pfn[0] = &print_i;
pfn[1] = &go_bye_bye;(pfn\[0\])();}
-- Russell Morris "So, broccoli, mother says you're good for me... but I'm afraid I'm no good for you!" - Stewy
Perfect Mr Russel Morris is the winner :-) BTW for the record, here's my code (very similar to Russel's as the basic concept of using func pointers is same)
void Dummy(int)
{
}void Show(int i);
typedef void (*FUNC)(int);
FUNC pFunc[2];void Show(int i)
{
printf("%d\t",i);
pFunc[i<100](i+1);
}int main()
{
pFunc[0] = &Dummy;
pFunc[1] = &Show;
Show(1);
} -
Perfect Mr Russel Morris is the winner :-) BTW for the record, here's my code (very similar to Russel's as the basic concept of using func pointers is same)
void Dummy(int)
{
}void Show(int i);
typedef void (*FUNC)(int);
FUNC pFunc[2];void Show(int i)
{
printf("%d\t",i);
pFunc[i<100](i+1);
}int main()
{
pFunc[0] = &Dummy;
pFunc[1] = &Show;
Show(1);
}In fact Russel's solution is better than mine. I used the < operator whoich is a conditional expression (it's valid usage as the question only disallows conditional statements) but Russel uses division which is perfect :-) Congrats again Russel. I raise my hat to you :-)
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Hmm...
static int i=1;
typedef void (*fn) ();
fn pfn[2];void print_i(void)
{
printf("%d\n",i);
// Not a conditional! :)
// i++ / 100 will evaluate to 0 until i has reached 100
(pfn[i++ / 100])();
}void go_bye_bye()
{
exit(0);
}void main(void)
{
pfn[0] = &print_i;
pfn[1] = &go_bye_bye;(pfn\[0\])();}
-- Russell Morris "So, broccoli, mother says you're good for me... but I'm afraid I'm no good for you!" - Stewy
Very nice. :) I hadn't even considered using function pointers. I was talking to Nish and suggested one printf with 100 digits and 100 linebreaks. ;)
[Cheshire] I can't afford those plastic things to cover the electric sockets so I just draw bunny faces on the electric outlets to scare the kids away from them... [RLtim] Newsflash! Kids aren't afraid of bunnies. [Cheshire] Oh they will be... -Bash.org
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Very nice. :) I hadn't even considered using function pointers. I was talking to Nish and suggested one printf with 100 digits and 100 linebreaks. ;)
[Cheshire] I can't afford those plastic things to cover the electric sockets so I just draw bunny faces on the electric outlets to scare the kids away from them... [RLtim] Newsflash! Kids aren't afraid of bunnies. [Cheshire] Oh they will be... -Bash.org
David Stone wrote: I was talking to Nish and suggested one printf with 100 digits and 100 linebreaks. Guys, it's okay, David is from Southern California - so as I said, it's quite okay :rolleyes:
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Hmm...
static int i=1;
typedef void (*fn) ();
fn pfn[2];void print_i(void)
{
printf("%d\n",i);
// Not a conditional! :)
// i++ / 100 will evaluate to 0 until i has reached 100
(pfn[i++ / 100])();
}void go_bye_bye()
{
exit(0);
}void main(void)
{
pfn[0] = &print_i;
pfn[1] = &go_bye_bye;(pfn\[0\])();}
-- Russell Morris "So, broccoli, mother says you're good for me... but I'm afraid I'm no good for you!" - Stewy
Mine was pretty similar except it doesn't use globals and uses a double-not rather than / 100 :)
#include <stdlib.h>
#include <stdio.h>typedef void (*funcptr)(int);
funcptr buf[2];
void count(int num)
{
printf("%d\n", 100-num);
(*buf[!!num])(num-1);
}int main()
{
buf[0] = exit;
buf[1] = count;
count(99);
}Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Can we print more than 100? Does the program have to terminate elegantly? I came up with a quick solution but it ain't exactly pretty :D cheers, Chris Maunder
Chris Maunder wrote: Can we print more than 100? Does the program have to terminate elegantly? I came up with a quick solution but it ain't exactly pretty :-D
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Mine was pretty similar except it doesn't use globals and uses a double-not rather than / 100 :)
#include <stdlib.h>
#include <stdio.h>typedef void (*funcptr)(int);
funcptr buf[2];
void count(int num)
{
printf("%d\n", 100-num);
(*buf[!!num])(num-1);
}int main()
{
buf[0] = exit;
buf[1] = count;
count(99);
}Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
*Applause* The double not was pretty cool - more elegant than Russel's division - and a lot more neater than my comparision operator :-) Nish
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*Applause* The double not was pretty cool - more elegant than Russel's division - and a lot more neater than my comparision operator :-) Nish
Ryan Russel still wins though - 1) he posted it first 2) he uses division which while inelegant is not a conditional expression, your double not (while elegant) is still a conditional expression (valid as the question only disallows conditional statements and expressions are not statements) Nish
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This question was asked in the Trivandrum Microsoft UG forums :- Using C, print 1-100 without using any loop or conditional statements? An answer was later posted in C++ (and the poster said that it was not possible without C++). Here's his C++ solution :-
static int i=0;
class ad
{
ad()
{
i++;
cout << i;
}
}void main()
{
ad objad[100];
}I spent a part of my Sunday evening wondering if it was possible in some alternative manner and found a solution. It shouldn't take you more than 15-20 minutes to think of it (assuming you haven't done this before). Let's see who gets there first (remember - no C++, no classes). Nish :-)
How about this
int main(int ac, char ** av) { static int i=1; printf("%d ",i++); i%101&&main(0,0); } -
How about this
int main(int ac, char ** av) { static int i=1; printf("%d ",i++); i%101&&main(0,0); }Cool stuff, Vivek :-) Nish
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Cool stuff, Vivek :-) Nish
Here's an improvement (LOC reduced anyway)
int main(int ac, char ** av)
{
printf("%d ",101-ac,(ac%100)&&main(ac+1,0));
}Run the app without arguments (first time ac will be 1) Nish
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BTW 100
printfs don't count :rolleyes: -
This question was asked in the Trivandrum Microsoft UG forums :- Using C, print 1-100 without using any loop or conditional statements? An answer was later posted in C++ (and the poster said that it was not possible without C++). Here's his C++ solution :-
static int i=0;
class ad
{
ad()
{
i++;
cout << i;
}
}void main()
{
ad objad[100];
}I spent a part of my Sunday evening wondering if it was possible in some alternative manner and found a solution. It shouldn't take you more than 15-20 minutes to think of it (assuming you haven't done this before). Let's see who gets there first (remember - no C++, no classes). Nish :-)
template <long val> void printfn() { std::cout << val << "\n"; printfn<val + 1>(); } template <> void printfn<100L>() { std::cout << "100\n"; } int main(int argc, char* argv[]) { printfn<1L>(); system("PAUSE"); return 0; }this works fine in VS.NET 2003 and gcc, but not with VC++ 6.0 which generates wrong specialization. i suppose that it also faster then previous solutions, since it is resolved in compile time. -
template <long val> void printfn() { std::cout << val << "\n"; printfn<val + 1>(); } template <> void printfn<100L>() { std::cout << "100\n"; } int main(int argc, char* argv[]) { printfn<1L>(); system("PAUSE"); return 0; }this works fine in VS.NET 2003 and gcc, but not with VC++ 6.0 which generates wrong specialization. i suppose that it also faster then previous solutions, since it is resolved in compile time.Good code, but C does not support templates :-) Nish
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Good code, but C does not support templates :-) Nish
:-O i got carried away by this line from your original post: "An answer was later posted in C++ (and the poster said that it was not possible without C++)."
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This question was asked in the Trivandrum Microsoft UG forums :- Using C, print 1-100 without using any loop or conditional statements? An answer was later posted in C++ (and the poster said that it was not possible without C++). Here's his C++ solution :-
static int i=0;
class ad
{
ad()
{
i++;
cout << i;
}
}void main()
{
ad objad[100];
}I spent a part of my Sunday evening wondering if it was possible in some alternative manner and found a solution. It shouldn't take you more than 15-20 minutes to think of it (assuming you haven't done this before). Let's see who gets there first (remember - no C++, no classes). Nish :-)
assuming that if also doesnt matter n yeah also mind that i m writing code directly here, neither compiled nor checked for output -
int main(int x) { printf("%d",x); if(x<100) main(++x); }program is expected to run from command prompt as filename 0 what do u think?? UTSAV MCA final yr -
This question was asked in the Trivandrum Microsoft UG forums :- Using C, print 1-100 without using any loop or conditional statements? An answer was later posted in C++ (and the poster said that it was not possible without C++). Here's his C++ solution :-
static int i=0;
class ad
{
ad()
{
i++;
cout << i;
}
}void main()
{
ad objad[100];
}I spent a part of my Sunday evening wondering if it was possible in some alternative manner and found a solution. It shouldn't take you more than 15-20 minutes to think of it (assuming you haven't done this before). Let's see who gets there first (remember - no C++, no classes). Nish :-)
Does this count? void main(){ char buffer[8] strcpy(buffer,"1-100"); cout << buffer << endl; } Jason