A Question!

**ZuK** · 01-13-2006

Either your book is wrong or you understood something wrong.
In c/c++ every non-zero value evaluates to true when used in a boolean context.

Kurt

**Blackroot** · 01-13-2006

so...

Code:

while(n) //is while(n >= 1)

So, != is just >=?

**vaibhav** · 01-13-2006

I don't understand why you guys are so much confused.
Just remember in while or if the statement under them is evaluated only if expression within them returns true

Code:

while()
{
statements;
}
False value-0
True value-Any non zero value

So,in above while loop the parenthesis can contain anything.
Following are all valid declaration.

Code:

while(0)//False condition means loop will not get executed
{}
while(1)//Always True means infinite loop
{}

So if you put a variable inside the parenthesis if the value of variable is 0 then it will get not get executed ,otherwise it will.

**hk_mp5kpdw** · 01-13-2006

Originally Posted by Blackroot

so...

Code:

while(n) //is while(n >= 1)

So, != is just >=?

!= is !=. >=1 does not include negative values which !=0 does include.

Code:

int n = -1;
while(n)
{
    cout << "Hello" << endl;
    ++n;
}

This will output "Hello" once.

So, while(n) is like saying while(n!=0). Get it?

**Salem** · 01-13-2006

> while(n) //is while(n >= 1)
Try it with a negative n and find out...

**vaibhav** · 01-13-2006

no,it is n!=0

Code:

int main()
{
int n=-1
while(n)
{
cout<<"Hello World";
n=0;
}
getch();
}

Above prog prints hello world once as expected.

**dwks** · 01-16-2006

Code:

int n=-1

Or some errors, like "missing semicolon", "undefined reference to cout", and ditto for getch().

Code:

while(!n) // Same as while(n==0)

You got that part right.

**siavoshkc** · 01-16-2006

Maybe some compilers work different but as far as I know any non zero value is true and zero is false. While(n) means stay in loop while n is true and exit if it is false. Lets take x=3 then in the program above we have this:
1)) i+1 so it will be 1.
2)) x-1 so x is 2.
3)) Comparing values, if one of (x or x-1) is non zero stay in loop, this is, so it stays in loop.
We should notice that value of x is not changed.
It stays in loop until x==0.

**siavoshkc** · 01-16-2006

4)) i+1 so it will be 2.
5)) x-1 so x is 1.
6)) x is 1 so stays in loop.
7)) i+1 so it will be 3.
8)) x-1 so x is 0.
9)) exits loop.
10)) prints out 3
As you can see, i is equal to first x (3).

And about !, it will turn true to untrue and vice versa.

**siavoshkc** · 01-16-2006

In the last loop x is 0 and x-- will be -1, huh? Yes, but after the comparison!
I think it is the difference between --x and x--. Funny isn't it?
If you replace x-- by --x then it never exits loop!
Because then it will first x-1, then determines T or F . So in all times it will be while(true).
In a||b if one of the operands are non zero it will be true.

**siavoshkc** · 01-16-2006

When we use (a||b), computer takes (a) (Left hand operand). If it is non-zero it will mark all (a||b) equal to true without examining b. If (a) was zero then it will examin b.
For this reason in (x||x--), x-- operation will not be executed until x==0. So only in last loop it will be executed(after "OR" operation of course) and if you add:

Code:

cout<<endl<<x;

x will be -1.

If we use --x, the operation will be executed in the last loop again but before "OR" operation!

If n is equal to zero, we have !n <=> n!=0
If n is non-zero, we have !n <=> n==0

**siavoshkc** · 01-16-2006

If we use x-- the code is equal to this

Code:

int x=5; int i=0;
do
{
	i++;
	x--;
}while(x);
x--;
cout<<i;

But if we use --x it is

Code:

int x=5; int i=0;
do
{
	i++;
	x--;
	if(x==0)x--;
}while(x);

cout<<i;

In second code, it wont exit the loop.

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