如何在Swift连接字符串?
在Objective-C中
NSString *string = @"Swift";
NSString *resultStr = [string stringByAppendingString:@" is a new Programming Language"];
or
NSString *resultStr=[NSString stringWithFormat:@"%@ is a new Programming Language",string];
但我想用swift语言来写。
当前回答
来自:Matt Neuburg书“iOS 13编程基础与Swift。”:
要组合(连接)两个字符串，最简单的方法是使用+运算符:
let s = "hello"
let s2 = " world"
let greeting = s + s2
这种方便的表示法是可能的，因为+操作符是重载的:当操作数是数字(数字加法)时它做一件事，当操作数是字符串(连接)时它做另一件事。 +操作符带有+=赋值快捷键;当然，左边的变量必须用var声明:
var s = "hello"
let s2 = " world"
s += s2
作为+=的替代方法，你可以调用append(_:)实例方法:
var s = "hello"
let s2 = " world"
s.append(s2)
连接字符串的另一种方法是使用joined(separator:)方法。你从一个要连接的字符串数组开始，并把要插入所有字符串之间的字符串交给它:
let s = "hello"
let s2 = "world"
let space = " "
let greeting = [s,s2].joined(separator:space)
其他回答
var language = "Swift"
var resultStr = "\(language) is a new programming language"
来自:Matt Neuburg书“iOS 13编程基础与Swift。”:
要组合(连接)两个字符串，最简单的方法是使用+运算符:
let s = "hello"
let s2 = " world"
let greeting = s + s2
这种方便的表示法是可能的，因为+操作符是重载的:当操作数是数字(数字加法)时它做一件事，当操作数是字符串(连接)时它做另一件事。 +操作符带有+=赋值快捷键;当然，左边的变量必须用var声明:
var s = "hello"
let s2 = " world"
s += s2
作为+=的替代方法，你可以调用append(_:)实例方法:
var s = "hello"
let s2 = " world"
s.append(s2)
连接字符串的另一种方法是使用joined(separator:)方法。你从一个要连接的字符串数组开始，并把要插入所有字符串之间的字符串交给它:
let s = "hello"
let s2 = "world"
let space = " "
let greeting = [s,s2].joined(separator:space)
\ this被用来将一个字符串附加到另一个字符串。
var first = "Hi"
var combineStr = "\(first) Start develop app for swift"
你也可以试试这个:- +关键字。
var first = "Hi"
var combineStr = "+(first) Start develop app for swift"
试试这段代码。
快速连接字符串
关于性能的几句话
UI测试包在iPhone 7(真实设备)，iOS 14， -Onone(调试，没有优化)[关于]
var result = ""
for i in 0...count {
<concat_operation>
}
计数= 5_000
//Append
result.append(String(i)) //0.007s 39.322kB
//Plus Equal
result += String(i) //0.006s 19.661kB
//Plus
result = result + String(i) //0.130s 36.045kB
//Interpolation
result = "\(result)\(i)" //0.164s 16.384kB
//NSString
result = NSString(format: "%@%i", result, i) //0.354s 108.142kB
//NSMutableString
result.append(String(i)) //0.008s 19.661kB
禁用下一个测试:
加上高达100_000 ~10s 插补可达100_000 ~10s NSString高达10_000 ->内存问题
Count = 1_000_000
//Append
result.append(String(i)) //0.566s 5894.979kB
//Plus Equal
result += String(i) //0.570s 5894.979kB
//NSMutableString
result.append(String(i)) //0.751s 5891.694kB
*注意将Int转换为字符串
源代码
import XCTest
class StringTests: XCTestCase {
let count = 1_000_000
let metrics: [XCTMetric] = [
XCTClockMetric(),
XCTMemoryMetric()
]
let measureOptions = XCTMeasureOptions.default
override func setUp() {
measureOptions.iterationCount = 5
}
func testAppend() {
var result = ""
measure(metrics: metrics, options: measureOptions) {
for i in 0...count {
result.append(String(i))
}
}
}
func testPlusEqual() {
var result = ""
measure(metrics: metrics, options: measureOptions) {
for i in 0...count {
result += String(i)
}
}
}
func testPlus() {
var result = ""
measure(metrics: metrics, options: measureOptions) {
for i in 0...count {
result = result + String(i)
}
}
}
func testInterpolation() {
var result = ""
measure(metrics: metrics, options: measureOptions) {
for i in 0...count {
result = "\(result)\(i)"
}
}
}
//Up to 10_000
func testNSString() {
var result: NSString = ""
measure(metrics: metrics, options: measureOptions) {
for i in 0...count {
result = NSString(format: "%@%i", result, i)
}
}
}
func testNSMutableString() {
let result = NSMutableString()
measure(metrics: metrics, options: measureOptions) {
for i in 0...count {
result.append(String(i))
}
}
}
}
在Swift 5中，apple引入了使用#符号的原始字符串。
例子:
print(#"My name is "XXX" and I'm "28"."#)
let name = "XXX"
print(#"My name is \#(name)."#)
符号#必须在\后面。常规的\(name)将被解释为字符串中的字符。
