can anyone crack this?

[PESITdude]

can anyone tell whats happening here?!

#include <iostream>

using namespace std;

int main()
{
	unsigned long a = 100000;
	long b = -1;
	if(b>a)
	cout<<"yes"<<endl;
	else
	cout<<"no"<<endl;

	return 0;
}

OUTPUT: yes

thanks in advance!

[Computergeek01]

Yeah "long" is short for "long int", int's don't do negative values instead they set the highest bit to '1' and process the rest of the number in reverse.

[Disch]

you're comparing signed and unsigned types.

What's happening is the compiler is casting to the "larger" type for the comparison. In this case it figures unsigned long is larger.

Therefore if(b>a) is the same as if((unsigned long)b>a)

(unsigned long)b casts a signed negative number (-1) to an indredibly large unsigned number (0xFFFFFFFF) which is larger than 100000

Hence it's printing "yes"

[PESITdude]

@Computergeek01

ok.. but how does that explain this..
a bit more detail pls...

also.. int stores negative numbers as well rite?
by default int means signed int

[PESITdude]

@Disch

Why does it do like that..
any logical reason to do like that?
i mean, why is this happenning>>>

(unsigned long)b casts a signed negative number (-1) to an indredibly large unsigned number (0xFFFFFFFF) which is larger than 100000

[chrisname]

Because you can't compare a negative integer to an unsigned integer. An unsigned integer can never be equal to -1 because it has no sign (no + or -; it is always positive). The only way to compare an unsigned integer with a signed integer is to cast one of them to the same type as the other one -- as Disch said, unsigned long is larger than long, so the compiler casts the long integer to an unsigned long integer. The value of (unsigned long)-1 is 0xFFFFFFFF. unsigned integer values wrap around, so -1 is equal to 0xFFFFFFFF, -2 is equal to FFFFFFFE, and so on.

[PESITdude]

@chrisname
woohoo!! understood... thanks pal :)

[chrisname]

You're welcome :]