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The expression 1==2 causes TypeScript to give the error Operator '==' cannot be applied to types '1' and '2'. What are the reasons for TypeScript treating these values as being of a different type (the typeof operator, rather predictably, says that they're both numbers)? Is it a design decision to specifically apply this to numbers, or a byproduct of the overall typing system? What harm could be caused by allowing parison of literals?

The expression 1==2 causes TypeScript to give the error Operator '==' cannot be applied to types '1' and '2'. What are the reasons for TypeScript treating these values as being of a different type (the typeof operator, rather predictably, says that they're both numbers)? Is it a design decision to specifically apply this to numbers, or a byproduct of the overall typing system? What harm could be caused by allowing parison of literals?

• javascript
• typescript
• types

Share Improve this question Follow asked Aug 31, 2017 at 11:27 clbclb 7231212 silver badges2323 bronze badges 8

• 1 thats just because the numbers have different values, it's not a type problem – kevinSpaceyIsKeyserSöze Commented Aug 31, 2017 at 11:29
• 3 But then surely the expression would just return false, rather than causing the script to not pile? – clb Commented Aug 31, 2017 at 11:33
• Well, it does pile, just with a warning… but yes, interesting question. Here's a link to reproduce: typescriptlang/play/#src=1%20%3D%3D%202%3B – deceze ♦ Commented Aug 31, 2017 at 11:33
• 2 @ZevSpitz "Why" is not really relevant to the question at hand. I'm asking about the design decision. – clb Commented Aug 31, 2017 at 11:35
• 2 @deceze there are no warnings, that is an error, but the process will still emit the js, unless the --noEmitOnError flag is used – Nitzan Tomer Commented Aug 31, 2017 at 11:36

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3 Answers 3

Sorted by: Reset to default 7

1 and 2 in this context are considered so called literal numbers. It means that value 1 has type 1 and thus can only be 1, same with 2. Given that, the expression of 1==2 doesn't make sense because 1 can never be 2, or more precisely their types mismatch, you can't pare apples to oranges.

Here are the rationale and in depth details on where literals types are assumed by default:

• https://github./Microsoft/TypeScript/pull/10676

About literal types:

• https://github./Microsoft/TypeScript/pull/9407
• https://github./Microsoft/TypeScript/pull/5185

One of many examples as to why literal types are useful:

• https://github./Microsoft/TypeScript/pull/9163

When TypeScript peforms type inference on the expression 1, it gives it the type 1, not the type number. You can see this if you examine code like this:

const a = 1;

If you use your IDE to query the inferred type of a, you'll see that the type of a is 1. In the TypeScript playground, for instance, you get a tooltip that says const a: 1.

So in if (1 == 2), 1 has type 1 and 2 has type 2. TypeScript does not allow you to pare them because they are of different inferred types. This is part of the type safety TypeScript gives you.

You can work around it with:

if (1 as number == 2) {
}

And you mentioned in a ment that you were doing the 1 == 2 parison because you could not do if (false) { ... } due to the piler plaining about unreachable code. I can work around that problem with this:

if (false as boolean) {
    console.log("something");
}

Typescript can create a type from any constant value. This when bined with union types creates a very powerful way of expressing what a function takes as an argument for example:

function doStuff(p : "yes"| 1| true| "no"| 0| false ){

}

doStuff("maybe"); //Error
doStuff(3); // Error
doStuff(1) ; //OK

You are experiencing the unfortunate side effect that errors such as yours instead of being reports as expression is always false turn into type patibility errors instead.

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