我可以在 Swift 中将范围运算符与 if 语句一起使用吗? [英] Can I use the range operator with if statement in Swift?
问题描述
是否可以将范围运算符 ...
和 ..<
与 if 语句一起使用.可能是这样的:
让 statusCode = 204如果 statusCode 在 200 ..<299 {NSLog("成功")}
您可以使用模式匹配"运算符 ~=
:
if 200 ... 299 ~= statusCode {打印(成功")}
或者带有表达式模式的 switch 语句(使用模式匹配内部操作符):
开关状态码{案例 200 ... 299:打印(成功")默认:打印(失败")}
请注意,..<
表示省略了上限值的范围,因此您可能想要200 ... 299
或 200 ...<300
.
附加信息: 当上面的代码在 Xcode 6.3 中编译时优化开启,然后进行测试
if 200 ... 299 ~= statusCode
实际上根本没有产生函数调用,只有三个汇编指令:
addq $-200, %rdicmpq 99 美元,%rdija LBB0_1
这是为
生成的完全相同的汇编代码if statusCode >= 200 &&状态代码 <= 299
你可以用
验证<前>xcrun -sdk macosx swiftc -O -emit-assembly main.swift从 Swift 2 开始,这可以写成
if case 200 ... 299 = statusCode {打印(成功")}
对 if 语句使用新引入的模式匹配.另请参阅 Swift 2 - if"中的模式匹配.
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")
}
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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