only one semicolon!!!

[sadi]

good day all,

I'm a new member
and I have a c++ challenge and need your help

the problem is

write a complete c++ program that read number and print it's square
note: you can use only one semicolon!!

I hope to have the answer as quickly as possible.

thanks for all

best regards

[Bazzy]

Only one semicolon in the entire program is quite difficult, the problem could mean that you should use << and >> stream operators.

If you really have to use just one semicolon try something like this:

#include <iostream>

#define sc ;
//sc will be replaced with a ';' by the preprocessor

int main()
{
    std::cout << "No semicolons here" sc
    return 0 sc
}

[sadi]

mmmmmmm
that to seem funny

thanks alot
if I use it correctly, the result should be like this

#include <iostream>

#define sc ;

int main ()

{
     int x sc

      std :: cout << "Enter number \n"  sc
      std :: cin >> x sc
      std :: cout << "The result = : " << x * x << std :: endl sc

      return 0 sc
}

isn't that

thanks alot for your help

regards ^_^

[rossman231]

This seems to be a pretty flawed challenge with many possible solutions.

I'd say the biggest issue is that they all depend on the program before pre-processing. iostream has plenty of semi-colons in it when expanded, as would the rest of your code. Another way that works on the same general loophole to do this is

// another file
#include <iostream>
using namespace std;

int cheat() {
  int x;
  cout << "x=" << endl;
  cin >> x;
  cout  << "x*x = " << x * x << endl;
  return 0;
}

// the "real" file
#include "file1.cpp" //should really be a header/cpp pair in real programming, but we don't need that here

int main(int argc, char** argv) {
  return cheat();
}

[sadi]

hhhh;

thank you very much,
good way !!

[rossman231]

Here's another way that is much less "cheaty," but requires C (not C++) for the lack of a return statement (and type of main).

#include <stdio.h>

main(int argc, char** argv) 
{
  int x, a = printf("x=\n"), b = scanf("%d", &x), c = printf("x*x=%d\n", x*x);
}

[helios]

It's legal for main to have a return type, but not a return statement (although it's generally advised that it has both).

[sadi]

Here's another way that is much less "cheaty," but requires C (not C++) for the lack of a return statement (and type of main).


#include <stdio.h>

main(int argc, char** argv)
{
int x, a = printf("x=\n"), b = scanf("%d", &x), c = printf("x*x=%d\n", x*x);
}

really nice
I like that
it will work in c++ if we write it like this

#include <stdio.h>

int main(int argc, char** argv) 
{
  int x, a = printf("x=\n"), b = scanf("%d", &x), c = printf("x*x=%d\n", x*x);
}

just add the function main type "int main"

good work
thanks for your help

I want to ask if you have any problems like this
I challenged my friend
and I want problems to ask him
can you help ??

regards

[SouravDutta]

I prefer to do it like this?

#include <stdio.h>
#include <stdlib.h>

int main(int argc,char*argv[])
{
      cout<<(gets(argv[1]),atoi(argv[1])*atoi(argv[1]));
}

No cheats, I suppose. Only the use of gets(), and argv[1] as a storage can raise eyebrows.. :)

[sadi]

anyway,
I have another solution
lets try something like this

#include <iostream>


void main()
{

int i;
if (std::cout<<"Enter the number:\n") {}
if (std::cin>>i) {}
if (std::cout<<"the result is: "<<i * i) {}

}

[writetonsharma]

hahahahaha... great minds on work.. :)