Converting (u)int64_t to NSNumbers

问题:

So essentially my question is this, I am creating an NSMutableDictionary using uint64_t objects as the key.

Is there any better way to create them than doing this?

uint64_t bob=7;

NSNumber *bobsNumber;

#if __LP64__ || TARGET_OS_EMBEDDED || TARGET_OS_IPHONE || TARGET_OS_WIN32 || NS_BUILD_32_LIKE_64
bobsNumber=[NSNumber numberWithUnsignedLong:bob];
#else
bobsNumber=[NSNumber numberWithUnsignedLongLong:bob];
#endif

This would work as long as you didn't include it in a binary file/sockets/NSData object/whatever. But is there any better way of doing this? I really would like to be sure that the object is 64-bits regardless of what platform I run it on.

I guess I could just avoid the whole issue by always going unsigned long long but of course that wastes tons of heap space on 64 bit machines if I allocate these objects in any significant numbers....

回答1:

long long is 64-bit on 64-bit OS X/iOS platforms. On all OpenStep-descended platforms, numberWithUnsignedLongLong: is correct for uint64_t.

Last time I checked, which factory method you use doesn’t actually affect the representation used anyway; it’s only dependent on the value of the number (unless you use a too-small size, causing it to be truncated).

Update: these days, the correct answer is NSNumber *bobsNumber = @(bob);.