How do you determine a number using only binary operators?

What I have to do is to cin in a variable, and tell what it is: positive or negative, and integer or floating point using binary operators only.





I did one with string, however, I have no idea how to do it with binary operators.





In a 32bit system, a number should be stored as 1 sign bit + 7 opponent bits + 24 mantissa. So I thought I could just shift the variable to the left by 31 times, I should get a 0 or 1, which tells me the sign of the number. But I get some weird numbers.





And i think it got something to do with the opponent bits to determine if it's integer or not. But the professor said it's got nothing to do with it. So I just need some hint to start with.





Here is a piece of code to show that I have done something, not just asking for the answer.








#include %26lt;cstdlib%26gt;


#include %26lt;iostream%26gt;


#include %26lt;string%26gt;





using namespace std;





void welcome();


void getLine(string %26amp;line);


void process(string %26amp;line);


void processLine(string %26amp;line);


bool again();





int main(int argc, char *argv[])


{


string line;





welcome();





do


{


getLine(line);


processLine(line);


}while(again());





system("PAUSE");


return EXIT_SUCCESS;


}





void welcome()


{


cout%26lt;%26lt;"This program identify the input as an integer or a double then negate the input"%26lt;%26lt;endl%26lt;%26lt;endl;


}





void getLine(string %26amp;line)


{


int redo=false;





do


{


int pos=0, dash=0, dot=0;


line.clear();


cout%26lt;%26lt;"Please enter a positive or a negative number without space: ";


getline(cin, line);


while((pos = line.find_first_not_of("-.0123456789")) != string::npos)


{


line.erase(pos,1);


}





for(int i=0; i%26lt;line.length(); i++)


{


if(line[i] == '-')


{


dash++;


}


if(line[i] == '.')


{


dot++;


}


}





if(dot%26gt;1 ||dash%26gt;1)


{


redo=true;


cout%26lt;%26lt;"You have entered more than 1 dot or 1 negative sign. "%26lt;%26lt;endl;


}





if(dash==1 %26amp;%26amp; line[0] != '-')


{


redo=true;


cout%26lt;%26lt;"The negative sign you entered is not at the front. "%26lt;%26lt;endl;


}





}while(redo==true);


}





void processLine(string %26amp;line)


{


int dash=0, dot=0;





for(int i=0; i%26lt;line.length(); i++)


{


if(line[i] == '-')


{


dash++;


}


if(line[i] == '.')


{


dot++;


}


}





cout%26lt;%26lt;"The number you entered is a ";





if(dash==0)


{


cout%26lt;%26lt;"positive ";


line.insert(0, "-");


}


else


{


cout%26lt;%26lt;"negative ";


line.erase(0, 1);


}





if(dot==0)


{


cout%26lt;%26lt;"integer. ";


}


else


{


cout%26lt;%26lt;"double. ";


}





cout%26lt;%26lt;endl%26lt;%26lt;"The negated number is : ";





for(int i=0; i%26lt;line.length(); i++)


{


cout%26lt;%26lt;line[i];


}





cout%26lt;%26lt;endl;


}





bool again()


{


char ans;





cout%26lt;%26lt;"Do you want to do this again? (Y)es/(N)o) ";


cin%26gt;%26gt;ans;


fflush(stdin);


cout%26lt;%26lt;endl;


return (toupper(ans) == 'Y');


}|||'%26lt;' and '%26gt;' are binary operators, I assume you're not allowed to say:

x %26gt; 0;

though, because that's the obvious solution.



A bad practice that might come in handy here would be to use the C function printf that's in stdio.h and use the %x or %p to print the actual hex representations of a few number to see what you're really working with.