CODEWITHSUNDEEP
iosobjective-ccmacros
wL_
wL_

Reputation: 1007

iPhone / iPad macro or c function?

What's the best way to define the following ternary operator?

[[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone ? x : y

I considered using the macro

#define phonePad(x, y) ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone ? x : y)

But this article mentions that it may not be the best idea. Is there a way to do the equivalent using a C function or is this the best way of implementing it?

Upvotes: 1

Views: 141

Answers (3)

mpontillo
mpontillo

Reputation: 13947

Neither.

BOOL isiPhone = ([[UIDevice currentDevice] userInterfaceIdiom] 
                 == UIUserInterfaceIdiomPhone);

foo = isiPhone ? x : y;
bar = isiPhone ? xprime : yprime;
...

If you turn that into a macro, you'll get a bunch of unnecessary calls into the Objective-C runtime. So just cache the result. Plus, it'll probably be a lot easier to read if you just write plain C code instead of using a macro.

Upvotes: 1

Mick MacCallum
Mick MacCallum

Reputation: 130193

I wouldn't use a macro for this. By using a macro, you require for the device to check the user interface idiom ever time this is used, and set x or y accordingly. Consider making a new method that returns based on the interface idiom. This can be static, because it's impossible for this value to ever change at runtime.

- (id)determineXOrY {
    static id obj = nil;

    static dispatch_once_t onceToken;
    dispatch_once(&onceToken, ^{
        obj = [[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone ? x : y
    });

    return obj;
}

Upvotes: 4

rmaddy
rmaddy

Reputation: 318794

If you do use the macro you must add some parentheses:

#define phonePad(x, y) ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone ? (x) : (y))

Without the parentheses you will have serious issues if x or y are more than simple values.

One downside to the macro is that you really need to be sure that both x and y evaluate to the same non-ambiguous data type.

Upvotes: 0

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