将“NSDecimalNumber”转换为“SInt64”而不会损失精度。 （SInt64、iOS 范围内）
目前，使用字符串 @"9999999999999999" 创建的 [NSDecimalNumber longLongValue] 返回 10000000000000000。
这意味着该类首先将其值转换为 double ，然后重新转换为 SInt64(signed long long)
如何避免这种行为？我想获得SInt64范围内的精确整数。
附言。 我考虑过使用 NSScanner 或 strtoll 转换为 NSString 并重新转换为 SInt64，但我相信有更好的选择方式。但如果你确定没有其他办法，请告诉我。
Currently, [NSDecimalNumber longLongValue] created with string @"9999999999999999" returns 10000000000000000.
This means the class converts it's value to double first, and re-converts into SInt64(signed long long)
How to evade this behavior? I want to get precise integral number within the range of SInt64.
PS.
I considered about converting to NSString and re-converting into SInt64 with NSScanner or strtoll, but I believe there's better way. But if you sure about there's no other way, please tell me that.
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首先：除非您确定它对性能至关重要，否则我会将其写入字符串并扫描回来。这是最简单的方法。
现在，如果您确实想这样做：
NSDecimalNumber获取一个NSDecimallong long尾数的值（可能引入检查来处理太大的尾数）First: unless you're sure it's performance-critical, I'd write it into a string and scan it back. That's the easy way.
Now, if you really want to do it otherwise:
NSDecimalfrom yourNSDecimalNumberlong longvalue from the mantissa (possibly introduce checks to handle too-large mantissas)从 NSDecimalNumber*originalValue 开始。
令 int64_t approx = [originalValue longLongValue]。这并不准确，但非常接近。
将approx转换为NSDecimalNumber，计算originalValue -approx，取longLongValue，并添加到approx。现在你得到了正确的结果。
Start with an NSDecimalNumber* originalValue.
Let int64_t approx = [originalValue longLongValue]. This will not be exact, but quite close.
Convert approx to NSDecimalNumber, calculate originalValue - approx, take the longLongValue, and add to approx. Now you got the correct result.