How to determine if compiling for 64-bit iOS in Xcode

Question

Consider the following function

CGSize CGSizeIntegral(CGSize size)
{
    return CGSizeMake(ceilf(size.width), ceilf(size.height));
}

CGSize actually consists of two CGFloats, and CGFloat's definition changes depending on the architecture:

typedef float CGFloat;// 32-bit
typedef double CGFloat;// 64-bit

So, the above code is wrong on 64-bit systems, and needs to be updated with something like

CGSize CGSizeIntegral(CGSize size)
{
#if 64_bit
    return CGSizeMake(ceil(size.width), ceil(size.height));
#else
    return CGSizeMake(ceilf(size.width), ceilf(size.height));
#endif
}

There is surely a compiler macro/constant for this (for Mac we can use INTEL_X86 for example) but I haven't been able to find this in the 64-bit transition guide.

How can I determine what architecture is being built for?

Answer 1

To determine if you are compiling for 64-bit, use __LP64__:

#if __LP64__
    return CGSizeMake(ceil(size.width), ceil(size.height));
#else
    return CGSizeMake(ceilf(size.width), ceilf(size.height));
#endif

__LP64__ stands for "longs and pointers are 64-bit" and is architecture-neutral.

According to your transition guide it applies for iOS as well:

The compiler defines the __LP64__ macro when compiling for the 64-bit runtime.

However, the preferred way to handle your use case is to use CGFLOAT_IS_DOUBLE. There is no guarantee that __LP64__ will always mean the CGFloat is a double, but it would be guaranteed with CGFLOAT_IS_DOUBLE.

#if CGFLOAT_IS_DOUBLE
    return CGSizeMake(ceil(size.width), ceil(size.height));
#else
    return CGSizeMake(ceilf(size.width), ceilf(size.height));
#endif