问题
I am looking at Xcode 7.3 notes and I notice this issue.
The ++ and -- operators have been deprecated
Could some one explain why it is deprecated? And am I right that in new version of Xcode now you going to use instead of ++ this x += 1;
Example:
for var index = 0; index < 3; index += 1 {
print(\"index is \\(index)\")
}
回答1:
A full explanation here from Chris Lattner, Swift's creator. I'll summarize the points:
- It's another function you have to learn while learning Swift
- Not much shorter than
x += 1 - Swift is not C. Shouldn't carry them over just to please C programmers
- Its main use is in C-style for loop:
for i = 0; i < n; i++ { ... }, which Swift has better alternatives, likefor i in 0..<n { ... }(C-style for loop is going out as well) - Can be tricky to read and maintain, for eg, what's the value of
x - ++xorfoo(++x, x++)? - Chris Lattner doesn't like it.
For those interested (and to avoid link rot), Lattner's reasons in his own words are:
These operators increase the burden to learn Swift as a first programming language - or any other case where you don't already know these operators from a different language.
Their expressive advantage is minimal - x++ is not much shorter than x += 1.
Swift already deviates from C in that the =, += and other assignment-like operations returns Void (for a number of reasons). These operators are inconsistent with that model.
Swift has powerful features that eliminate many of the common reasons you'd use ++i in a C-style for loop in other languages, so these are relatively infrequently used in well-written Swift code. These features include the for-in loop, ranges, enumerate, map, etc.
Code that actually uses the result value of these operators is often confusing and subtle to a reader/maintainer of code. They encourage "overly tricky" code which may be cute, but difficult to understand.
While Swift has well defined order of evaluation, any code that depended on it (like foo(++a, a++)) would be undesirable even if it was well-defined.
These operators are applicable to relatively few types: integer and floating point scalars, and iterator-like concepts. They do not apply to complex numbers, matrices, etc.
Finally, these fail the metric of "if we didn't already have these, would we add them to Swift 3?"
回答2:
I realize that this comment doesn't answer the question nevertheless there may be people looking for a solution how to keep these operators working and such a solution can be found in the bottom. 😇
I personally prefer ++ and -- operators. I can't agree with the opinion that they are tricky or hard to manage. Once the developer understand what these operators do (and we are talking about pretty simple stuff) the code should be very clear.
In the explanation why the operators were deprecated is mentioned that their main use was in C-style for loops. I don't know about others but I personally don't use C-style loops at all and there are still many other places or situations when ++ or -- operator is useful.
I would like to also mention that varName++ returns a value so it can be used in the return whereas varName += 1 can not.
For any of you who would like to keep these operators working here is the solution:
prefix operator ++ {}
postfix operator ++ {}
prefix operator -- {}
postfix operator -- {}
// Increment
prefix func ++(inout x: Int) -> Int {
x += 1
return x
}
postfix func ++(inout x: Int) -> Int {
x += 1
return (x - 1)
}
prefix func ++(inout x: UInt) -> UInt {
x += 1
return x
}
postfix func ++(inout x: UInt) -> UInt {
x += 1
return (x - 1)
}
prefix func ++(inout x: Int8) -> Int8 {
x += 1
return x
}
postfix func ++(inout x: Int8) -> Int8 {
x += 1
return (x - 1)
}
prefix func ++(inout x: UInt8) -> UInt8 {
x += 1
return x
}
postfix func ++(inout x: UInt8) -> UInt8 {
x += 1
return (x - 1)
}
prefix func ++(inout x: Int16) -> Int16 {
x += 1
return x
}
postfix func ++(inout x: Int16) -> Int16 {
x += 1
return (x - 1)
}
prefix func ++(inout x: UInt16) -> UInt16 {
x += 1
return x
}
postfix func ++(inout x: UInt16) -> UInt16 {
x += 1
return (x - 1)
}
prefix func ++(inout x: Int32) -> Int32 {
x += 1
return x
}
postfix func ++(inout x: Int32) -> Int32 {
x += 1
return (x - 1)
}
prefix func ++(inout x: UInt32) -> UInt32 {
x += 1
return x
}
postfix func ++(inout x: UInt32) -> UInt32 {
x += 1
return (x - 1)
}
prefix func ++(inout x: Int64) -> Int64 {
x += 1
return x
}
postfix func ++(inout x: Int64) -> Int64 {
x += 1
return (x - 1)
}
prefix func ++(inout x: UInt64) -> UInt64 {
x += 1
return x
}
postfix func ++(inout x: UInt64) -> UInt64 {
x += 1
return (x - 1)
}
prefix func ++(inout x: Double) -> Double {
x += 1
return x
}
postfix func ++(inout x: Double) -> Double {
x += 1
return (x - 1)
}
prefix func ++(inout x: Float) -> Float {
x += 1
return x
}
postfix func ++(inout x: Float) -> Float {
x += 1
return (x - 1)
}
prefix func ++(inout x: Float80) -> Float80 {
x += 1
return x
}
postfix func ++(inout x: Float80) -> Float80 {
x += 1
return (x - 1)
}
prefix func ++<T : _Incrementable>(inout i: T) -> T {
i = i.successor()
return i
}
postfix func ++<T : _Incrementable>(inout i: T) -> T {
let y = i
i = i.successor()
return y
}
// Decrement
prefix func --(inout x: Int) -> Int {
x -= 1
return x
}
postfix func --(inout x: Int) -> Int {
x -= 1
return (x + 1)
}
prefix func --(inout x: UInt) -> UInt {
x -= 1
return x
}
postfix func --(inout x: UInt) -> UInt {
x -= 1
return (x + 1)
}
prefix func --(inout x: Int8) -> Int8 {
x -= 1
return x
}
postfix func --(inout x: Int8) -> Int8 {
x -= 1
return (x + 1)
}
prefix func --(inout x: UInt8) -> UInt8 {
x -= 1
return x
}
postfix func --(inout x: UInt8) -> UInt8 {
x -= 1
return (x + 1)
}
prefix func --(inout x: Int16) -> Int16 {
x -= 1
return x
}
postfix func --(inout x: Int16) -> Int16 {
x -= 1
return (x + 1)
}
prefix func --(inout x: UInt16) -> UInt16 {
x -= 1
return x
}
postfix func --(inout x: UInt16) -> UInt16 {
x -= 1
return (x + 1)
}
prefix func --(inout x: Int32) -> Int32 {
x -= 1
return x
}
postfix func --(inout x: Int32) -> Int32 {
x -= 1
return (x + 1)
}
prefix func --(inout x: UInt32) -> UInt32 {
x -= 1
return x
}
postfix func --(inout x: UInt32) -> UInt32 {
x -= 1
return (x + 1)
}
prefix func --(inout x: Int64) -> Int64 {
x -= 1
return x
}
postfix func --(inout x: Int64) -> Int64 {
x -= 1
return (x + 1)
}
prefix func --(inout x: UInt64) -> UInt64 {
x -= 1
return x
}
postfix func --(inout x: UInt64) -> UInt64 {
x -= 1
return (x + 1)
}
prefix func --(inout x: Double) -> Double {
x -= 1
return x
}
postfix func --(inout x: Double) -> Double {
x -= 1
return (x + 1)
}
prefix func --(inout x: Float) -> Float {
x -= 1
return x
}
postfix func --(inout x: Float) -> Float {
x -= 1
return (x + 1)
}
prefix func --(inout x: Float80) -> Float80 {
x -= 1
return x
}
postfix func --(inout x: Float80) -> Float80 {
x -= 1
return (x + 1)
}
prefix func --<T : BidirectionalIndexType>(inout i: T) -> T {
i = i.predecessor()
return i
}
postfix func --<T : BidirectionalIndexType>(inout i: T) -> T {
let y = i
i = i.predecessor()
return y
}
回答3:
Apple has removed the ++ and made it much simpler with the another old traditional way.
Instead of ++, you need to write +=.
Example:
var x = 1
//Increment
x += 1 //Means x = x + 1
Similarly for decrement operator --, you need to write -=
Example:
var x = 1
//Decrement
x -= 1 //Means x = x - 1
For for loops:
Increment Example:
Instead of
for var index = 0; index < 3; index ++ {
print("index is \(index)")
}
You can write:
//Example 1
for index in 0..<3 {
print("index is \(index)")
}
//Example 2
for index in 0..<someArray.count {
print("index is \(index)")
}
//Example 3
for index in 0...(someArray.count - 1) {
print("index is \(index)")
}
Decrement Example:
for var index = 3; index >= 0; --index {
print(index)
}
You can write:
for index in 3.stride(to: 1, by: -1) {
print(index)
}
//prints 3, 2
for index in 3.stride(through: 1, by: -1) {
print(index)
}
//prints 3, 2, 1
for index in (0 ..< 3).reverse() {
print(index)
}
for index in (0 ... 3).reverse() {
print(index)
}
Hope this helps!
回答4:
Chris Lattner has gone to war against ++ and --. He writes, “Code that actually uses the result value of these operators is often confusing and subtle to a reader/maintainer of code. They encourage “overly tricky” code which may be cute, but difficult to understand….While Swift has well defined order of evaluation, any code that depended on it (like foo(++a, a++)) would be undesirable even if it was well-defined…these fail the metric of “if we didn’t already have these, would we add them to Swift 3?””
Apple wanted to keep swift a clean, clear, non-confusing and straight-to-the-point language. And so they deprecated ++ and -- keyword.
回答5:
The Fix-it feature of Xcode gives clear answer to this.
Replace ++ increment operator with old-fashioned value += 1 (short-hand operator) and -- decrement operator with value -= 1
回答6:
From the docs:
The increment/decrement operators in Swift were added very early in the development of Swift, as a carry-over from C. These were added without much consideration, and haven't been thought about much since then. This document provides a fresh look at them, and ultimately recommends we just remove them entirely, since they are confusing and not carrying their weight.
回答7:
For Swift 4, you can restore the ++ and -- operators as extensions for Int and other types. Here is an example:
extension Int{
static prefix func ++(x: inout Int) -> Int {
x += 1
return x
}
static postfix func ++(x: inout Int) -> Int {
defer {x += 1}
return x
}
static prefix func --(x: inout Int) -> Int {
x -= 1
return x
}
static postfix func --(x: inout Int) -> Int {
defer {x -= 1}
return x
}
}
It works the same way for other types, such as UIInt, Int8, Float, Double, etc.
You can paste these extensions in a single file in your root directory, and they will be available for use inside all of your other files there.
I've noticed a couple of down-votes for my answer here, almost as soon as I posted it. Which I take as a philosophical disagreement, rather than criticism of how my code works. It works perfectly, if you check it out in a playground.
The reason why I posted this answer is because I disagree with making computer programming languages unnecessarily different from each other.
Having a lot of similarities between languages makes them easer for people to learn and to switch from one language to another.
Developers normally use several programming languages, rather than just one. And it's a real hassle to switch from one language to another, when there are no conventions and no common standardization across languages.
I believe that there should be syntax differences between languages only as much as necessary, and not any more than this.
回答8:
Here is a generic version of some of the code posted so far. I would voice the same concerns as others: it is a best practice to not use these in Swift. I agree that this could be confusing for those reading your code in the future.
prefix operator ++
prefix operator --
prefix func ++<T: Numeric> (_ val: inout T) -> T {
val += 1
return val
}
prefix func --<T: Numeric> (_ val: inout T) -> T {
val -= 1
return val
}
postfix operator ++
postfix operator --
postfix func ++<T: Numeric> (_ val: inout T) -> T {
defer { val += 1 }
return val
}
postfix func --<T: Numeric> (_ val: inout T) -> T {
defer { val -= 1 }
return val
}
This can also be written as an extension on the Numeric type.
回答9:
var value : Int = 1
func theOldElegantWay() -> Int{
return value++
}
func theNewFashionWay() -> Int{
let temp = value
value += 1
return temp
}
This is definitely a downside, right?
回答10:
In Swift 4.1 it could be achieved this way:
prefix operator ++
postfix operator ++
extension Int{
static prefix func ++(x: inout Int)->Int{
x += 1
return x
}
static postfix func ++(x: inout Int)->Int{
x += 1
return x-1
}
}
//example:
var t = 5
var s = t++
print("\(t) \(s)")
Notice that despite the fact that this solution in similar to previous solutions in this post, they don't work anymore in Swift 4.1 and this example does.
Also notice that whomever above mentions that += is a replacement for ++ just don't fully understand the operator as ++ combined with assignment is actually two operations, hence a shortcut.
In my example: var s = t++ does two things: assign the value of t to s and then increment t. If the ++ comes before, it's the same two operations done in reversed order.
To my opinion, the reasoning of Apple about why to remove this operator(mentioned in previous answers), is not only false reasoning but furthermore I believe it is a lie and the true reason is that they couldn't make their compiler handle it. It gave them troubles in previous versions so they gave up.
The logic of "too complicated to understand operator, hence removed" is obviously a lie because Swift contains operators far more complicated and much less useful which were not removed. Also, the vast majority of programming languages has it.
JavaScript, C, C#, Java, C++ and so many more. Programmers happily use it.
Whomever it is too difficult to understand this operator for, they and only they should do the += (or perhaps s = s + 1 if += is too complexed as well).
The strategy behind Swift is simple: Apple believes the programmer is dumb and therefore should be treated accordingly.
The truth is that Swift, launched at September 2014 was supposed to be somewhere else by now. Other languages grew up much faster.
I can list many major mistakes in the language, from serious ones: such as arrays pasted by value and not by reference, to annoying ones: variadic parameters functions can't accept an array which is the whole idea behind it. I don't think that Apple's employees are even allowed to look at other languages such as Java so they don't even know that Apple is light years behind. Apple could have adopted Java as a language but these days, challenge is not technology, but ego is. If they would have opened IntelliJ to write some Java, they would for sure close their business understanding that at this point, they can't and won't catch up ever.
来源：https://stackoverflow.com/questions/35158422/the-and-operators-have-been-deprecated-xcode-7-3