Is it possible to use the range operator ... and ..< with if statement. Maye something like this:
let statusCode = 204
if statusCode in 200 ..< 299 {
NSLog("Success")
}
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6 Answers
You can use the "pattern-match" operator ~=:
if 200 ... 299 ~= statusCode {
print("success")
}
Or a switch-statement with an expression pattern (which uses the pattern-match operator internally):
switch statusCode {
case 200 ... 299:
print("success")
default:
print("failure")
}
Note that ..< denotes a range that omits the upper value, so you probably want
200 ... 299 or 200 ..< 300.
Additional information: When the above code is compiled in Xcode 6.3 with optimizations switch on, then for the test
if 200 ... 299 ~= statusCode
actually no function call is generated at all, only three assembly instruction:
addq $-200, %rdi
cmpq $99, %rdi
ja LBB0_1
this is exactly the same assembly code that is generated for
if statusCode >= 200 && statusCode <= 299
You can verify that with
xcrun -sdk macosx swiftc -O -emit-assembly main.swift
As of Swift 2, this can be written as
if case 200 ... 299 = statusCode {
print("success")
}
using the newly introduced pattern-matching for if-statements. See also Swift 2 - Pattern matching in "if".
13 Comments
Sky
Cool, is this O(1)? Also, it'd be nice if Swift had a short hand for switch statements, like Scala for example. But given that you're always forced to handle all possibilities at compile time in Swift, it may not really be feasible.
Martin R
@Sky: From the generated assembly code one can see that a library function
func ~= (Range<A>, A) -> Bool is called. I would assume that this function works with O(1).
codester
@MartinR how you get to know which function called by assembly language.Hopper? +1 for cool answer
Martin R
@codester: I compiled the code on the command line with
xcrun -sdk macosx swift -emit-assembly main.swift and inspected the assembly code. Then I used xcrun swift-demangle ... to de-mangle the name of the called function. - Unfortunately, Xcode cannot yet create assembly code for Swift files, perhaps it will work in a later version.Martin R
@DuncanC: developer.apple.com/library/ios/documentation/Swift/Conceptual/…, section "Expression Pattern": "... By default, the ~= operator compares two values of the same type using the == operator. It can also match an integer value with a range of integers in an Range object, ..."
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This version seems to be more readable than pattern matching:
if (200 ... 299).contains(statusCode) {
print("Success")
}
2 Comments
Nazim Kerimbekov
Exactly what I was searching for
Ahmadreza
I get this error => Can't form Range with upperBound < lowerBound
This is an old thread, but it seems to me we're over-thinking this. It seems to me the best answer is just
if statusCode >= 200 && statusCode <= 299
There's no
if 200 > statusCode > 299
form that I'm aware of, and the other suggested solutions are doing function calls, which are harder to read, and might be slower to execute. The pattern match method is a useful trick to know, but seems like a poor fit for this problem.
Edit:
Personally, I find the pattern match operator to be hideous, and wish the compiler would support if x in 1...100 syntax. That is sooooo much more intuitive and easy to read than if 1...100 ~= x
7 Comments
Martin R
You are right that this version is better to read, I just tried to answer the explicit question "Is it possible to use the range operator ... ?" – But Xcode 6.3 beta (in optimized mode) generates exactly three assembly instructions for
if 200 ... 299 ~= statusCode, no function call :)Martin R
Actually
if 200 ... 299 ~= statusCode gives the same assembly code as if statusCode >= 200 && statusCode <= 299
rickster
Unless this conditional is in a critical section that gets visited thousands of times per second, worrying about function call overhead is premature optimization. Even then, I'd worry more about what a function call is doing rather than the cost of calling it. Nice job @MartinR for proving there's no cost regardless, though.
Duncan C
@rickster, true enough. I still tend to prefer efficient constructs over inefficient ones as a matter of habit (assuming readability is similar). Not to the extent that I waste too much of MY time on it, but it still pays to know what the costs of different approaches are.
RenniePet
This is nitpicking, but I disagree with your suggestion that your if statement is more readable or understandable than the answer posted by @SerhiiYakovenko. Simply on the basis of DRY - you name "statusCode" twice. In a late-night bleary-eyed debugging session after I'd decided that a different variable named "statusValue" should be used here instead of "statusCode", I just might make the mistake of changing one of the variable names and not the other.
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I wanted to check 4xx errors except 401. Here is the code:
let i = 401
if 400..<500 ~= i, i != 401 {
print("yes")
} else {
print("NO")
}
Comments
I preferred Range .contains() operator too, until found that its implementation is inefficient - https://oleb.net/blog/2015/09/swift-ranges-and-intervals/
We can represent the condition x < 0 using a range: (Int.min..<0).contains(x) is exactly equivalent. It is vastly slower, though. The default implementation of contains(_:) traverses the entire collection, and executing a loop nine quintillion times in the worst case is not cheap.
1 Comment
KPM
This information is obsolete, and wrong. It already was wrong at the time you wrote your answer.
if you want to know if the response status code is a success, simply make it this way
if response.statuscode < 300 {
print("response is a success")
}
if you iterate from 200,201,203,204 ... 300. There were just too much unnecessary iterations. Hope this helps! :D
