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")
}
Types of Range in Swift In Swift, there are three types of range: Closed Range. Half-Open Range. One-Sided Range.
Ranges in Swift allow us to select parts of Strings, collections, and other types. They're the Swift variant of NSRange which we know from Objective-C although they're not exactly the same in usage, as I'll explain in this blog post. Ranges allow us to write elegant Swift code by making use of the range operator.
operator. Returns a Boolean value indicating whether a value is included in a range.
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".
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