{"id":59664,"date":"2018-05-02T16:43:06","date_gmt":"2018-05-02T11:13:06","guid":{"rendered":"https:\/\/www.vskills.in\/certification\/tutorial\/?p=59664"},"modified":"2024-04-12T14:13:22","modified_gmt":"2024-04-12T08:43:22","slug":"typescript-type-inference","status":"publish","type":"page","link":"https:\/\/www.vskills.in\/certification\/tutorial\/typescript-type-inference\/","title":{"rendered":"TypeScript Type Inference"},"content":{"rendered":"

TypeScript Type Inference<\/h1>\n

In TypeScript, there are several places where type inference is used to provide type information when there is no explicit type annotation. For example, in this code<\/p>\n

let x = 3;<\/p>\n

The type of the x variable is inferred to be number. This kind of inference takes place when initializing variables and members, setting parameter default values, and determining function return types.<\/p>\n

In most cases, type inference is straightforward. In the following sections, we\u2019ll explore some of the nuances in how types are inferred.
\nBest common type<\/p>\n

When a type inference is made from several expressions, the types of those expressions are used to calculate a \u201cbest common type\u201d. For example,<\/p>\n

let x = [0, 1, null];<\/p>\n

To infer the type of x in the example above, we must consider the type of each array element. Here we are given two choices for the type of the array: number and null. The best common type algorithm considers each candidate type, and picks the type that is compatible with all the other candidates.<\/p>\n

Because the best common type has to be chosen from the provided candidate types, there are some cases where types share a common structure, but no one type is the super type of all candidate types. For example:<\/p>\n

let zoo = [new Rhino(), new Elephant(), new Snake()];<\/p>\n

Ideally, we may want zoo to be inferred as an Animal[], but because there is no object that is strictly of type Animal in the array, we make no inference about the array element type. To correct this, instead explicitly provide the type when no one type is a super type of all other candidates:<\/p>\n

let zoo: Animal[] = [new Rhino(), new Elephant(), new Snake()];<\/p>\n

When no best common type is found, the resulting inference is the union array type, (Rhino | Elephant | Snake)[].<\/p>\n

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