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IOS/Objective-C/Swift/Speech:声明SFSpeech Recognizer变量时指定区域设置

sfspeechrecognizer objective-c ios

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user6631314 · 技术社区 · 6 年前

Swift显然允许您在声明变量时指定speechRecognizer的区域设置,如下所示:

private let speechRecognizer = SFSpeechRecognizer(locale: Locale.init(identifier: "en-US"))

在Objective-C中可以这样做吗?现在我在接口中声明了一个变量:

SFSpeechRecognizer *speechRecognizer;

speechRecognizer = [[SFSpeechRecognizer alloc] initWithLocale:[[NSLocale alloc] initWithLocaleIdentifier:@"en-US"]];

理想情况下,我想在声明的一开始就这样做,但我不清楚Swift和Objective-C到底在做什么。

谢谢你的建议和见解。

1 回复 | 直到 6 年前

1

CodeBender 6 年前

想象一下,Swift调用的结构是这样的:

// Create a Locale object for US English
let locale = Locale.init(identifier: "en-US")
// Create a speech recognizer object for US English
let speechRecognizer = SFSpeechRecognizer(locale: locale)

然后将Swift代码与目标C进行比较:

// Here you are create an uninitialized variable of type SFSpeechRecognizer
// this will then hold the SFSpeechRecognizer when you initialize it in the next line
SFSpeechRecognizer *speechRecognizer;
// This is accomplishing the same logic as the above Swift call
speechRecognizer = [[SFSpeechRecognizer alloc] initWithLocale:[[NSLocale alloc] initWithLocaleIdentifier:@"en-US"]];

如果您希望将objective-c调用改写为一行,则可以将其改为这样:

SFSpeechRecognizer *speechRecognizer = [[SFSpeechRecognizer alloc] initWithLocale:[[NSLocale alloc] initWithLocaleIdentifier:@"en-US"]];

这两种方法都没有错,只是Swift可以推断变量类型,所以在启动语音识别器之前不需要创建空变量。目标C can 推断变量类型,因此命令可能被拆分,只是为了使行稍微短一点。