Question

我有代码

let z;
z = 50;
z = 'z';

我的tsconfig.json是：

{
  "compilerOptions": {
    "target": "es5",
    "module": "commonjs",
    "sourceMap": false,
    "noEmitOnError": true,
    "strict": true,
    "noImplicitAny": true
  }
}

但是到底有什么例外可以编译到js？

最诚挚的问候， Crova

Answer 1

因为永远不会将z输入为any。 z的类型只是根据您分配给它的内容推断出来。

来自release notes：

使用TypeScript 2.1，而不是只选择任何，TypeScript将根据您最后分配的内容推断类型。

示例：

let x; // You can still assign anything you want to 'x'. x = () => 42; // After that last assignment, TypeScript 2.1 knows that 'x' has type '() => number'. let y = x(); // Thanks to that, it will now tell you that you can't add a number to a function! console.log(x + y); // ~~~~~ // Error! Operator '+' cannot be applied to types '() => number' and 'number'. // TypeScript still allows you to assign anything you want to 'x'. x = "Hello world!"; // But now it also knows that 'x' is a 'string'! x.toLowerCase();

所以在你的情况下：

let z; z = 50; let y = z * 10; // `z` is number here. No error z = 'z'; z.replace("z", "")// `z` is string here. No error

Answer 2

noImplicitAny字面意思是：

如果TypeScript使用＆＃39;任何＆＃39;触发错误无论何时它无法推断a 型

在上面的例子中，代码编译器的任何一点都可以轻松推断出z的类型。因此，它可以检查您是否允许在z上调用适当的方法/道具。

TS编译 - ＆＃34; noImplicitAny＆＃34;没有工作

2 个答案: