为什么在TypeScript中连接字符串和未定义的变量不会引发错误

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英文:

Why concatenating string and undefined does Not raise an error in TypeScript

问题

例如,这段代码无效:

function fn1(a: number, b?: number) {
  console.log(a + b);  // error 'b' is possibly 'undefined'.(18048)
}
fn1(5);

但这段代码是有效的:

function fn2(a: string, b?: string) {
  console.log(a + b);  // valid code
}
fn2('Hi');

为什么在严格模式下 TypeScript 在第二种情况下没有引发错误?

英文:

For example, this code is not valid:

function fn1(a: number, b?: number) {
  console.log(a + b);  // error 'b' is possibly 'undefined'.(18048)
}
fn1(5);

But this code is valid:

function fn2(a: string, b?: string) {
  console.log(a + b);  // valid code
}
fn2('Hi');

Why doesn't Typescript in strict mode raise an error in the second case?

答案1

得分: 1

这个问题的权威答案可以在 Microsoft TypeScript 团队的开发负责人在 microsoft/TypeScript#2093 的一个评论中找到:

在很多代码中都有这样的操作:

console.log("Foo is " + x);

其中 x 可能是一个函数,undefinednull,在这种情况下显式调用 toString() 将不是一个好主意。我们决定,向字符串添加任何东西是如此常见,以至于它永远不会是一个错误。

[这个建议] 不符合引入语言中的破坏性更改的标准。

所以答案就是:不管你是否同意,TypeScript 团队不想破坏现有代码,其中会将各种东西连接到字符串上。


在该问题以及类似的问题(例如 microsoft/TypeScript#7989microsoft/TypeScript#42983microsoft/TypeScript#48962)中提到,那些真正希望看到这个功能并愿意使用 ESLint 的人可以启用 the @typescript-eslint/restrict-plus-operands 规则,该规则将在使用 + 连接不同类型的值时发出错误。因此,即使它们不是 TypeScript 的直接组成部分,仍然有选择。

英文:

The authoritative answer to this question can be found in a comment on microsoft/TypeScript#2093 by the development lead of the TypeScript team at Microsoft:

> There's a huge amount of code out there that does stuff like this:
>
> console.log("Foo is " + x);
>
> where x might be a function, undefined or null, in which case explicitly calling toString() would be a bad idea. We made the decision that adding anything to a string is so often done that it is not ever an error.
>
> [This suggestion] doesn't meet the bar for introducing a breaking change in the language.

So that's the answer: whether or not you agree with it, the TypeScript team didn't want to break existing real-world code that concatenates random stuff to strings.


In that issue and similar issues (such as microsoft/TypeScript#7989, microsoft/TypeScript#42983, and microsoft/TypeScript#48962), it is mentioned that those who really want to see this functionality and are willing to use ESLint could enable the @typescript-eslint/restrict-plus-operands rule which will issue errors when using + to concatenate values of different types. So there are options out there, even if they are not part of TypeScript directly.

答案2

得分: 0

因为在拼接中,如果第一个值是string,JavaScript会强制第二个值也是string,这就是为什么编译器不关心b可能是未定义的,因为你可以将任何东西与string拼接。

然而,如果你尝试将a的类型设为number,或进行任何其他操作,就会出现错误:

function fn2(a: number, b?: string) {
  console.log(a + b);  // 'b' is possibly 'undefined'
}
fn2(5);
英文:

Because in concatenation, if the first value is string JavaScript will force the second value to be string that is why the compiler does not care if b may be undefined, since you can concatenate anything with a string.

However, if you try to make a of type number, or try any other operation the error will occur :

function fn2(a: number, b?: string) {
  console.log(a + b);  // 'b' is possibly 'undefined'
}
fn2(5);

答案3

得分: 0

我有不同的看法

我认为字符串 + 未定义是可以接受的,因为这是一个有效的JS操作

但数字不能与未定义连接,让我感到惊讶,因为这也是一个有效的JS操作

英文:

I have a different thought

I think string + undefined is acceptable because it is a valid JS operation

but number cannot concatenate undefined surprise me because it is also a valid JS operation

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  • 本文由 发表于 2023年2月18日 11:35:05
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