将（u）int64_t转换为NSNumbers [英] Converting (u)int64_t to NSNumbers

问题描述

因此，本质上我的问题是这是，我创建一个NSMutableDictionary使用uint64_t对象作为关键。

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

有没有更好的方式来创建它们？ / p>

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

这将工作，只要你没有包括在二进制文件/套接字/ NSData对象/任何。但是有没有更好的方法这样做？我真的想确保对象是64位，而不管我运行它的平台。

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.

我想我可以避免整个问题，总是去unsigned long long但当然，如果我分配这些对象在任何有效的数字，64位机器上浪费吨堆空间....

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....

推荐答案

在64位OS X / iOS平台上，

long long 为64位。在所有OpenStep下降的平台上， numberWithUnsignedLongLong：适用于 uint64_t 。

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

上次我检查，你使用哪个工厂方法实际上不影响使用的表示;它只依赖于数字的值（除非你使用太小的大小，导致它被截断）。

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).

更新：这些天，正确的答案是 NSNumber * bobsNumber = @（bob）; 。

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

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