Is this C statement valid?

NO

1

if( ( x == CONDITION1 ) ||

2

    ( x == CONDITION2 ) )

Having said that:
It is valid in the sense, that it follows the language rules. But it 
does not what you think it should do.

if ( (x == CONDITION1) || (x == CONDITION2) )
{

    //stuff
}


The 'stuff' section will be processed if one of the conditions is true.

this

1

if(x == (CONDITION1 || CONDITION2)) {

2

3

    //stuff

4

}

is equal to this

1

if((x == CONDITION1) || (x == CONDITION2)) {

2

3

    //stuff

4

}

(only) when

1

BOOL x = YES;     // must be YES

2

3

BOOL CONDITION1 = ...;

4

BOOL CONDITION2 = ...;

So, to answer your question "Is this C statement valid?":
Yes it is.
But it works differently like you expect.
The language C don't understand booleans.
It takes numbers to represent true or false.

false == 0;
true  != 0;  // common 1

Your statement:

1

if(x == (CONDITION1 || CONDITION2))

2

{

3

4

    //stuff

5

}

checks first the statement within the parenthesizes.
It returns a 1 when CONDITION1 OR CONDITION2 is not 0.
Otherwise it returns a 0.

Then it checks x with the return value.
When x is 0 and both conditions are also 0 the equation returns a 
1.
When x is 1 and at least one condition is not 0 the equation returns 
a 1, too.

When the expression in the if-header is not 0 the if branch will be 
executed. Otherwise not, of course.

All this does not hurt the rules of C.

____________________________

But you purpose was to check if x has the same value like CONDITION1 
OR CONDITION2.
To do this you have to reorder your question.
Has x the same value like CONDITION1 OR has x the same value like 
CONDITION2?

Therefore is the answer of kbuchegg and the others correct.

1

if( ( x == CONDITION1 ) ||  ( x == CONDITION2 ) )

2

{

3

    //stuff

4

}

This expression returns a 1 when at least one of the statements within 
the parenthesizes return a 1.

regards
BrEin




*************************************
I hope this answer is not "overdressed" and I seam to be a squirt.
I thought the question was caused by missing some knowledge.
Sorry if not.

Fabian Hoemcke write:
> What is right?
> a 1?
> or
> an 1?
>
> I serios don't know it? :D

Since 1 ("one") doesn't start with a spoken vowel, it's "a 1" and not
"an 1". But in this context

> ... the equation returns a 1.

I think it's better style to omit the article:

  ... the equation returns 1.

Klaus Wachtler write:
> there is only one 1
> :-)

I have at least three of them. Would you like to see them?

        1          1          1    ;-)

SNCR

#define cond(x) (CONDITION1==x || CONDITION2==x)

by chance?