#if defined (x) || (y) ; is this valid?

During one review I came across a piece of code like the following :

#if defined(x) || y 

What does the above statement mean ? Will the condition execute properly ?

4 Answers

Yes it is valid.

Here is what the Standard (C99) says in 6.10p1:

if-group: # if constant-expression new-line groupopt # ifdef identifier new-line groupopt # ifndef identifier new-line groupopt 

the defined operator is seen as unary operator part of a constant expression (6.10.1p1).

In your example, the condition is evaluated as true if the macro x is defined OR if y is defined and different than 0

The reasoning for this is twofold.

Instead of using a #ifdef, you use the defined operator so that you can use logical operators on it (&&, ||, etc.), so that you don't have to duplicate your code so that it is included properly if there is multiple criteria for what you need to defined.

Also, in my opinion, I find it much easier to read as #if defined(x) than #ifdef x, and you could do the following #if defined(x) && defined(y), whereas that isn't possible with #ifdef.

Yes, since defined(x) is a boolean and returns true or false.

The above statement means "either x is defined or y is true".

it is correct but it is a bad practice. Why use two different evaluations ('y' and 'defined(x)') in one conditional. It might be confusing for others. The example below shows more common use of || and && operatiors against #defined macros:

#define AA 1 #define BB 2 #if (defined AA) || (defined BB) #warning "A or B" #endif #if (defined AA) && (defined BB) #warning "A and B" #endif 

When the above code was run user would get these messages on screen:

#warning "A or B" #warning "A and B" 

But if code was like this (AA was undefined):

#undefine AA #define BB 2 #if (defined AA) || (defined BB) #warning "A or B" #endif #if (defined AA) && (defined BB) #warning "A and B" #endif 

then user will get this message:

#warning "A or B"