typescript 打字稿中的“不可分配给类型为 never 的参数”错误是什么？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/52423842/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
What is "not assignable to parameter of type never" error in typescript?
提问by Lev
Code is:
代码是：
const foo = (foo: string) => {
const result = []
result.push(foo)
}
I get the following TS error:
我收到以下 TS 错误：
[ts] Argument of type 'string' is not assignable to parameter of type 'never'.
[ts] 'string' 类型的参数不能分配给 'never' 类型的参数。
What am I doing wrong? Is this a bug?
我究竟做错了什么？这是一个错误吗？
回答by Tha'er M. Al-Ajlouni
All you have to do is define your resultas a string array, like the following:
您所要做的就是将您定义result为一个字符串数组，如下所示：
const result : string[] = [];
Without defining the array type, it by default will be never. So when you tried to add a string to it, it was a type mismatch, and so it threw the error you saw.
没有定义数组类型，默认情况下它将是never. 所以当你试图向它添加一个字符串时，它是一个类型不匹配，所以它抛出了你看到的错误。
回答by neomib
Another way is :
另一种方法是：
const result = [] as any;
回答by Randy Hudson
This seems to be a recent regression or some strange behavior in typescript. If you have the code:
这似乎是最近的回归或打字稿中的一些奇怪行为。如果你有代码：
const result = []
Usually it would be treated as if you wrote:
通常它会被当作你写的：
const result:any[] = []
however, if you have both noImplicitAnyFALSE, ANDstrictNullChecksTRUE in your tsconfig, it is treated as:
但是，如果您的 tsconfig 中同时具有noImplicitAnyFALSE和strictNullChecksTRUE，则将其视为：
const result:never[] = []
This behavior defies all logic, IMHO. Turning on null checks changes the entry types of an array?? And then turning on noImplicitAnyactually restores the use of anywithout any warnings??
这种行为违反了所有逻辑，恕我直言。打开空检查会更改数组的条目类型？？然后打开noImplicitAny居然恢复使用any没有任何警告？？
When you truly have an array of any, you shouldn't need to indicate it with extra code.
当你真的有一个 的数组时any，你不需要用额外的代码来指示它。
回答by ImFarhad
I was having same error In ReactJSstatless function while using ReactJs Hook useState. I wanted to set state of an object array , so if I use the following way
在使用 ReactJs Hook useState时，我在ReactJSstatless 函数中 遇到了同样的错误。我想设置对象数组的状态，所以如果我使用以下方式
const [items , setItems] = useState([]);
and update the state like this:
并像这样更新状态：
const item = { id : new Date().getTime() , text : 'New Text' };
setItems([ item , ...items ]);
I was getting error:
Argument of type '{ id: number; text: any }' is not assignable to parameter of type 'never'
我收到错误：
'{ id: number; 类型的参数; 文本：任何 }' 都不能分配给类型为 'never' 的参数
but if do it like this,
但如果这样做，
const [items , setItems] = useState([{}]);
Error is gone but there is an item at 0 indexwhich don't have any data(don't want that).
错误消失了，但索引为0的项目没有任何数据（不想要）。
so the solution I found is:
所以我找到的解决方案是：
const [items , setItems] = useState([] as any);
回答by neiya
I got the same error in ReactJS function component, using ReactJS useState hook. The solution was to declare the type of useState at initialisation:
我在 ReactJS 函数组件中遇到了同样的错误，使用 ReactJS useState 钩子。解决方案是在初始化时声明 useState 的类型：
const [items , setItems] = useState<IItem[]>([]); // replace IItem[] with your own typing: string, boolean...
回答by Mickers
I was able to get past this by using the Array keyword instead of empty brackets:
我能够通过使用 Array 关键字而不是空括号来解决这个问题：
const enhancers: Array<any> = [];
Use:
用：
if (typeof devToolsExtension === 'function') {
enhancers.push(devToolsExtension())
}
回答by domready
You need to type resultto an array of string const result: string[] = [];.
您需要键入resultstring 数组const result: string[] = [];。
回答by Lakshan Hettiarachchi
The solution i found was
我找到的解决方案是
const [files, setFiles] = useState([] as any);
回答by Ashutosh Singh
Remove "strictNullChecks": truefrom "compilerOptions"or set it to false in the tsconfig.jsonfile of your Ng app. These errors will go away like anything and your app would compile successfully.
删除真：“strictNullChecks”从“compilerOptions”或将其设置为false在tsconfig.json您伍应用程序文件。这些错误会像任何事情一样消失，您的应用程序将成功编译。
Disclaimer: This is just a workaround. This error appears only when the null checks are not handled properly which in any case is not a good way to get things done.
免责声明：这只是一种解决方法。仅当未正确处理空检查时才会出现此错误，这在任何情况下都不是完成任务的好方法。
