在 TypeScript 中，如何将布尔值转换为数字，例如 0 或 1
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In TypeScript, How to cast boolean to number, like 0 or 1
提问by a2htray yuen
As we know, the type cast is called assertion type in TypeScript. And the following code section:
众所周知，类型转换在 TypeScript 中称为断言类型。以及以下代码部分：
// the variable will change to true at onetime
let isPlay: boolean = false;
let actions: string[] = ['stop', 'play'];
let action: string = actions[<number> isPlay];
On compiling, it go wrong
编译的时候出错
Error:(56, 35) TS2352: Neither type 'boolean' nor type 'number' is assignable to the other.
Then I try to use the anytype:
然后我尝试使用any类型：
let action: string = actions[<number> <any> isPlay];
Also go wrong. How can I rewrite those code.
也会出错。我怎样才能重写这些代码。
回答by Nitzan Tomer
You can't just cast it, the problem is at runtime not only at compile time.
你不能只是投射它，问题不仅在编译时出现在运行时。
You have a few ways of doing that:
你有几种方法可以做到这一点：
let action: string = actions[isPlay ? 1 : 0];
let action: string = actions[+isPlay];
let action: string = actions[Number(isPlay)];
Those should be fine with both the compiler and in runtime.
这些对于编译器和运行时都应该没问题。
回答by user239558
You can convert anything to boolean and then to a number by using +!!:
您可以使用以下命令将任何内容转换为布尔值，然后转换为数字+!!：
const action: string = actions[+!!isPlay]
This can be useful when for example you want at least two out of three conditions to hold, or exactly one to hold:
例如，当您希望至少满足三个条件中的两个，或正好满足一个条件时，这会很有用：
const ok = (+!!something) + (+!!somethingelse) + (+!!thirdthing) > 1
const ok = (+!!something) + (+!!somethingelse) + (+!!thirdthing) === 1
