x == (y || z)

[Dipset]

if (x == (y || z))

Is there a way to get something like that besides

if (x == y || x == z)

[daat99]

Originally Posted by Dipset

if (x == (y || z))

Is there a way to get something like that besides

if (x == y || x == z)

Not in c#, sorry.

[yarex]

[x & (y || z)] || [!x & (!y & !z)]

[Phantom]

Whats wrong with?

if (x == y || x == z)

What you want to do, and what you asked, does exactly the samething so I perhaps don't understand why the method that does work is not something you want to use.

[wieganka]

Originally Posted by Phantom

Whats wrong with?

if (x == y || x == z)

What you want to do, and what you asked, does exactly the samething so I perhaps don't understand why the method that does work is not something you want to use.

I believe that this is a question of being more "efficient" (or lazy, whichever you prefer). The poster wanted to know if there was a shorter way of writing:

if (x == y || x == z)

Which, of course, in C#, there isn't.

[Phantom]

Originally Posted by wieganka

I believe that this is a question of being more "efficient" (or lazy, whichever you prefer). The poster wanted to know if there was a shorter way of writing:

if (x == y || x == z)

Which, of course, in C#, there isn't.

I don't understand whats wrong with it....

[wieganka]

There's absolutely nothing wrong with it. They were just looking for a shorter way of writing it.

[Dipset]

if (x == y || x == z) looks fucking ugly and is a waste of text.

[Phantom]

Originally Posted by Dipset

if (x == y || x == z) looks fucking ugly and is a waste of text.

If you say so, but considering there isn't another way to do an OR statement, its not a waste of text.

[Nochte]

Originally Posted by Dipset

if (x == y || x == z) looks fucking ugly and is a waste of text.

you're kidding, right?

[wieganka]

Originally Posted by Dipset

if (x == y || x == z) looks fucking ugly and is a waste of text.

I don't even want to get in on this one....sheesh...

[yarex]

(x == y || x == z) is wrong.

x == (y || z) .... when x ==0 and y ==1 and z==0
0 == (1||0) .... = (0==1) = 0. so it gives false

but when its (x == y || x == z)

(0 == 1 || 0==0) ....... (0 || 1) = 1

So these two are not equal in case if x=0 and y=1 and z=0
or case
x=0 and y=0 and x=1
(other cases are ok )

[Jeff]

Originally Posted by Dipset

if (x == y || x == z) looks fucking ugly and is a waste of text.

That accually made me laugh out loud irl hahaha great.

[Sep102]

Originally Posted by yarex

(x == y || x == z) is wrong.

x == (y || z) .... when x ==0 and y ==1 and z==0
0 == (1||0) .... = (0==1) = 0. so it gives false

but when its (x == y || x == z)

(0 == 1 || 0==0) ....... (0 || 1) = 1

So these two are not equal in case if x=0 and y=1 and z=0
or case
x=0 and y=0 and x=1
(other cases are ok )

Yes, but in his case, the semantics of

x == y || x == z

is what he was looking for, just in a shorter way of doing it, so the semantics that you pointed out, while valid in certain situations, was not what he was looking for.


Now, back to the topic, you can doing something like

x == (y || z)

But not normally and not all too pretty (nor without abusing the privilege of overloading operators). I felt like screwing with C#'s overloading of operators and wanted to see if you could abuse it in somewhat the same way as in C++, and I'm happy to say that you can, to a degree.

class TestClass
{
	public TestClass ( int i1, int i2 )
	{
		x = i1;
		y = i2;
	}
		
	public static bool operator == ( int i, TestClass tc )
	{
		return (i == tc.x || i == tc.y);
	}

	public static bool operator != ( int i, TestClass tc )
	{
		return !(i == tc);
	}

	int x
	        , y;
}

class TestClass2
{
	public TestClass2 ( int i )
	{
		x = i;
	}

	public static TestClass operator | ( TestClass2 tc1, TestClass2 tc2 )
	{
		return new TestClass( tc1.x, tc2.x );
	}
	
	public int x;
}

Using these two classes together, it's possible for me to write something of the form

int i = 20;
TestClass2 tc1 = new TestClass2(10)
        , tc2 = new TestClass2(20);

i == (tc1 | tc2);

and have it be semantically equivalent to

i == tc1.x || i == tc2.x

Please note that I do not advocate doing this, as it is indeed a bit of a misuse of the powerful and easily abusable operator overloading concept.

However, do note that I love that you can do this.

[David]

Originally Posted by Dipset

if (x == y || x == z) looks fucking ugly and is a waste of text.

As opposed to what, this thread maybe?

[FingersMcSteal]

Originally Posted by Dipset

if (x == y || x == z) looks fucking ugly and is a waste of text.

Best answer i've read in a long time

[Phantom]

Originally Posted by yarex

(x == y || x == z) is wrong.

x == (y || z) .... when x ==0 and y ==1 and z==0
0 == (1||0) .... = (0==1) = 0. so it gives false

but when its (x == y || x == z)

(0 == 1 || 0==0) ....... (0 || 1) = 1

So these two are not equal in case if x=0 and y=1 and z=0
or case
x=0 and y=0 and x=1
(other cases are ok )

Your example its actually wrong, because 0 == 0 is true, and thus the if statement is true.

What he wants is if x is equal to y or z, then the statement would be true, which is what has been posted.

[Dipset]

Originally Posted by David

As opposed to what, this thread maybe?

Yes, I asked a simple question and everyone goes on and on about random shit.

This shouldn't have gone past daat99's reply.

[Phantom]

Originally Posted by Dipset

Yes, I asked a simple question and everyone goes on and on about random shit.

This shouldn't have gone past daat99's reply.

I just wanted to know why he didn't want to use it.

Now that I know, I wish I didn't because its a lame reason

[yarex]

Originally Posted by Phantom

Your example its actually wrong, because 0 == 0 is true, and thus the if statement is true.

What he wants is if x is equal to y or z, then the statement would be true, which is what has been posted.

If you mean this:
(0 == 1 || 0==0) ....... (0 || 1) = 1

then think again, becouse you are wrong.

0==1 gives 0
0==0 gives 1
therefore i've written that its equal to (0 || 1) = 1 = true

Result is the same as you say, but looks like you think something different.

[Phantom]

I don't understand what your saying about what I said is incorrect.

You said the following:
0 == (1||0) .... = (0==1) = 0. so it gives false

Looking at it again, I see thats actually is infact true, you do understand that || stands for OR correct? If the OR statement was an bit-wsie OR statement then it would be false. But its not and 0==0 is a true statement.

[Sep102]

Perhaps this would clarify his statement a bit better

We have:

false == (true || false);

Now, first we evaluate the (true || false):

(true || false) = true

Next we evaluate the rest of the statement, substituting in the first part we evaluated:

false == (true || false) = false == true = false

So, therefore, in his statement he is correct in that it does indeed evaluate to false.

[Dipset]

You guys take things a bit to far.

[Phantom]

Originally Posted by Sep102

Perhaps this would clarify his statement a bit better

We have:

false == (true || false);

Now, first we evaluate the (true || false):

(true || false) = true

Next we evaluate the rest of the statement, substituting in the first part we evaluated:

false == (true || false) = false == true = false

So, therefore, in his statement he is correct in that it does indeed evaluate to false.

See the problem with your the statement its "false" is beause the statement itself is incorrect. Your doing an operation that is illegal in a sense. Taking the statement into context, what the real value is, is what the correct statement would do.

I am still going to say your incorrect for saying its false.

[Ravatar]

	switch(x)
	{
		case y:
		case z:
		{
		
		} break;
	}

:P