[DemonEdge]

there was a loop in the midterm (c++) like this:

x = 0;
y= 0;
while(!x)
{
etc...etc...
}
etc....



what is important to me is what da **** does not X mean?
i think y != x would sound much understandable but !x alone i don't understand...
can some1 help me plz?

[intercodes]

DemonEdge,

while (!x) means while (1) which means the while condition is true and evaluates the while function.

[See Plus Plus]

!x is any non-zero number.

[DemonEdge]

thx for the fast replies..
hmmm...
you say that while(!x) means while(1)
is that because the prof. declared the x = 0 @ the beginning?
and does that mean that the loop will run for ever?

[intercodes]

DemonEdge,

yes, aslong as the while function has an integer value of '1' , the while condition is satisfied. So this loop runs infinitly, unless the 'etc' is replaced with some code.

Any value other than 1 in the 'while' will exit from the function.

[DemonEdge]

k the whole program is as follows:
( of course i didn't even manage to solve it)
**********************************************
#include <iostream.h>
void main()
{
int x = 0;
int y = 0;
while(!x)
{
switch(x)
{
case 0:
for(x=3;x>0;x--)
{
switch(y)
{
case 0:
cout<<"yes"<<endl;
break;
default:
cout<<"no"<<endl;
break;
}
}
if(y>0)
x = y;
break;
case 1:
break;
case 2:
breal;
}
y++;
}
}
*******************************************
so...can some1 explain this program to me in detail plz??

[intercodes]

DemonEdge,

int x = 0;
int y = 0;
while(!x)
{

'while' condition is true, and the function is evaluated..so it follows...

while(!x)
{
switch(x)
{
case 0:
for(x=3;x>0;x--)
{

Now, here at the switch statement ,value of x is 0. So switch is executed i.e case 0 condition. ..

for(x=3;x>0;x--)
{
switch(y)
{
case 0:
cout<<"yes"<<endl;
break;
default:
cout<<"no"<<endl;
break;
}
}

For 3 loops the above block will be executed [value of y is 0 , switch(y) satisfies].Produces output "yes" "yes" "yes"

if(y>0)
x = y;
break;
case 1:
break;
case 2:
breal;
}
y++;
}
}

none of the above statement will be excuted except for y++ .so y=1 now.

The while loop continues , and enters the switch(y) fn where case 'default' is satisified ,so prints "no" no" "no".

y>0 satisifies, x=y is done , the while loop is terminated as while(!x) is false

[Iron_Cross]

#include <iostream.h>
void main()
{
    int x = 0;         // Initialize the vars to zero
    int y = 0;
    // the NOT operator just means the oposite...since 0 can mean false and 1 can be mean true
    // He's esentially saying "If x means true, make it mean false...if x means false, make it mean true"
    // In this case, it means true, because it originally meant false
    while(!x)          
    {
	// Now he's trying to figure out, what x was originally
        switch(x)
        {
	    // If x was originally 0 (which, in this particular case it was do this...
            case 0:
		// He's now initializing x to 3 and decrementing it until it reaches 1
		// esentially that is 3 times. And since y is never changed, it will 
		// output "yes" to the screen 3 times.
               for(x=3;x>0;x--)
               {
		     // he now wants to find out what y equals
                     switch(y)
                     {
			  // in this case y (look at top of page) equals 0 so the first one will output
                          case 0:
			      // "yes" will be output
                              cout<<"yes"<<endl;
                              break;
                          default:
                              cout<<"no"<<endl;
                              break;
                     }
               }
		// since y is NOT greater than 0, x does not get changed
               if(y>0)
                   x = y;
               break;
	    // this fails because x = 0;
            case 1:
                break;
	    // same here, x does not = 2;
            case 2:
                breal;
        }
	// y now equals 1
        y++;
    }
}

You can take the rest from there..that should give you some basic idea about how the code operates.