I'm migrating a project from Swift 2.2 to Swift 3 , and I'm trying to get rid of old Cocoa data types when possible.
My problem is here: migrating NSDecimalNumber
to Decimal
.
I used to bridge NSDecimalNumber
to Double
both ways in Swift 2.2 :
let double = 3.14
let decimalNumber = NSDecimalNumber(value: double)
let doubleFromDecimal = decimalNumber.doubleValue
Now, switching to Swift 3 :
let double = 3.14
let decimal = Decimal(double)
let doubleFromDecimal = ???
decimal.doubleValue
does not exist, nor Double(decimal)
, not even decimal as Double
... The only hack I come up with is:
let doubleFromDecimal = (decimal as NSDecimalNumber).doubleValue
But that would be completely stupid to try to get rid of NSDecimalNumber
, and have to use it once in a while...
Well, either I missed something obvious, and I beg your pardon for wasting your time, or there's a loophole needed to be addressed, in my opinion...
Thanks in advance for your help.
Edit : Nothing more on the subject on Swift 4 .
Edit : Nothing more on the subject on Swift 5 .
NSDecimalNumber
and Decimal
are bridged
The Swift overlay to the Foundation framework provides the Decimal structure, which bridges to the NSDecimalNumber class. The Decimal value type offers the same functionality as the NSDecimalNumber reference type, and the two can be used interchangeably in Swift code that interacts with Objective-C APIs. This behavior is similar to how Swift bridges standard string, numeric, and collection types to their corresponding Foundation classes. Apple Docs
but as with some other bridged types certain elements are missing.
To regain the functionality you could write an extension:
extension Decimal {
var doubleValue:Double {
return NSDecimalNumber(decimal:self).doubleValue
}
}
// implementation
let d = Decimal(floatLiteral: 10.65)
d.doubleValue
Another solution that works in Swift 3 is to cast the Decimal
to NSNumber
and create the Double
from that.
let someDouble = Double(someDecimal as NSNumber)
As of Swift 4.2 you need:
let someDouble = Double(truncating: someDecimal as NSNumber)
Solution that works in Swift 4
let double = 3.14
let decimal = Decimal(double)
let doubleFromDecimal = NSDecimalNumber(decimal: decimal).doubleValue
print(doubleFromDecimal)
let doubleValue = Double(truncating: decimalValue as NSNumber)
Decimal
in Swift 3 is not NSDecimalNumber
. It's NSDecimal
, completely different type.
You should just keep using NSDecimalNumber
as you did before.
You are supposed to use as
operator to cast a Swift type to its bridged underlying Objective-C type. So just use as
like this.
let p = Decimal(1)
let q = (p as NSDecimalNumber).doubleValue
In Swift 4, Decimal
is NSDecimalNumber
. Here's citation from Apple's official documentation in Xcode 10.
Important
The Swift overlay to the Foundation framework provides the
Decimal
structure, which bridges to theNSDecimalNumber
class. For more information about value types, see Working with Cocoa Frameworks in Using Swift with Cocoa and Objective-C (Swift 4.1).
There's no NSDecimal
anymore. There was confusing NSDecimal
type in Swift 3, but it seems to be a bug. No more confusion.
I see the OP is not interested in Swift 4, but I added this answer because mentioning only about (outdated) Swift 3 made me confused.
In Swift open source, the implementation is actually done in Decimal.swift
, but it is private. You can re-use the code from there.
extension Double {
@inlinable init(_ other: Decimal) {
if other._length == 0 {
self.init(other._isNegative == 1 ? Double.nan : 0)
return
}
var d: Double = 0.0
for idx in (0..<min(other._length, 8)).reversed() {
var m: Double
switch idx {
case 0: m = Double(other._mantissa.0)
break
case 1: m = Double(other._mantissa.1)
break
case 2: m = Double(other._mantissa.2)
break
case 3: m = Double(other._mantissa.3)
break
case 4: m = Double(other._mantissa.4)
break
case 5: m = Double(other._mantissa.5)
break
case 6: m = Double(other._mantissa.6)
break
case 7: m = Double(other._mantissa.7)
break
default: break
}
d = d * 65536 + m
}
if other._exponent < 0 {
for _ in other._exponent..<0 {
d /= 10.0
}
} else {
for _ in 0..<other._exponent {
d *= 10.0
}
}
self.init(other._isNegative != 0 ? -d : d)
}
}
For swift 5, the function is
let doubleValue = Double(truncating: decimalValue as NSNumber)
the example in the below, show the number of float.
let decimalValue: Decimal = 3.14159
let doubleValue = Double(truncating: decimalValue as NSNumber)
print(String(format: "%.3f", doubleValue)) // 3.142
print(String(format: "%.4f", doubleValue)) // 3.1416
print(String(format: "%.5f", doubleValue)) // 3.14159
print(String(format: "%.6f", doubleValue)) // 3.141590
print(String(format: "%.7f", doubleValue)) // 3.1415900
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