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A static type system can help prevent many potential runtime errors as applications grow, which is why Vue 3 is written in TypeScript. This means you don't need any additional tooling to use TypeScript with Vue - it has a first-class citizen support.
//this enables stricter inference for data properties on `this`
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//これは `this` のデータプロパティに対してより厳密な推論を可能にします
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"strict":true,
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"jsx":"preserve",
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"moduleResolution":"node"
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}
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}
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```
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Note that you have to include `strict: true` (or at least`noImplicitThis: true` which is a part of `strict` flag) to leverage type checking of `this`in component methods otherwise it is always treated as `any`type.
For developing Vue applications with TypeScript, we strongly recommend using [Visual Studio Code](https://code.visualstudio.com/), which provides great out-of-the-box support for TypeScript. If you are using [single-file components](./single-file-components.html) (SFCs), get the awesome [Vetur extension](https://github.com/vuejs/vetur), which provides TypeScript inference inside SFCs and many other great features.
TypeScript should be able to infer most of the types without defining types explicitly. For example, if you have a component with a number `count`property, you will have an error if you try to call a string-specific method on it:
const result =this.count.split('') // => Property 'split' does not exist on type 'number'
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const result =this.count.split('') // => 'split' プロパティは 'number' 型に存在しません
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}
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})
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```
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If you have a complex type or interface, you can cast it using [type assertion](https://www.typescriptlang.org/docs/handbook/basic-types.html#type-assertions):
Because of the circular nature of Vue’s declaration files, TypeScript may have difficulties inferring the types of computed. For this reason, you may need to annotate the return type computed properties.
If you find validator not getting type inference or member completion isn’t working, annotating the argument with the expected type may help address these problems.
const result =year.value.split('') // => Property 'split' does not exist on type 'number'
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const result =year.value.split('') // => 'split' プロパティは 'number' 型に存在しません
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}
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})
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```
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Sometimes we may need to specify complex types for a ref's inner value. We can do that simply passing a generic argument when calling ref to override the default inference:
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