Is it possible to create a typeguard for a complex combinatory type while retaining the existing type information?

Question

Is it possible to create a typeguard for a complex combinatory type while retaining the existing type information?

Is there a method to create a TypeScript typeguard for a complex combinatory type that includes multiple "and" and "or" statements to check for the presence of one of the "or" types?

For example:

interface Type1 { cat: string }
    interface Type2 { dog: boolean }
    interface Type3 { mouse: number }
    interface Type4 { elephant: string | number }
    type CombinatoryType = (Type1 & (Type2 | Type3)) | Type4;

If we have typeguards to verify the existence of Type2 and Type4 (named hasType2 and hasType4 respectively), calling them in a specific order should yield certain results:

function doSomething(data: CombinatoryType) {
        if (hasType2(data)) {
            // 'data' now has type: (Type1 & Type2) | Type4

            if (hasType4(data)) {
                // 'data' is now of type: Type1 & Type2 & Type4
           }
        }
    }

And calling them in reverse order should produce different outcomes:

function doSomething(data: CombinatoryType) {
        if (hasType4(data)) {
            // 'data' is now of type: (Type1 & (Type2 | Type3)) & Type4

            if (hasType2(data)) {
                // 'data' now holds type: Type1 & Type2 & Type4
           }
        }
    }

Key points to consider:

The typeguard must retain previous type information without losing it due to prior assertions.
Avoid setting an explicit return type like data is (Type1 & Type2) | Type4 as the combinatory type could become extensive, potentially causing issues with VSCode's type inference (although intellisense remains functional).

javascript typescript typeguards

Answer 1

Answer №1

If my interpretation is accurate based on the code you provided, it seems like you are aiming to establish type guards for both Type2 and Type4 within the CombinatoryType type. Am I correct in this assumption?

If that's the case, here is an illustration for you:

function hasType2(data: CombinatoryType): data is (Type1 & Type2) | Type4 {
   return 'dog' in data;
}

function hasType4(data: CombinatoryType): data is (Type1 & (Type2 | Type3)) & Type4 {
   return 'elephant' in data;
}

In the snippet above, we utilize the in operator to verify the existence of the dog property in data for hasType2, and similarly for the elephant property in data for hasType4.

The in operator acts as a type guard by confirming whether a property exists within an object.

The syntax data is (Type1 & Type2) | Type4 and data is (Type1 & (Type2 | Type3)) & Type4 serves to inform TypeScript that if the function returns true, then data conforms to the specified type.

These type guards can now be applied within your doSomething function:

function doSomething(data: CombinatoryType) {
   if (hasType2(data)) {
       // 'data' now falls under the type: (Type1 & Type2) | Type4

       if (hasType4(data)) {
           // 'data' now fits the mold of Type1 & Type2 & Type4
       }
   }
}

In this scenario, if hasType2(data) yields true, then within the initial if, data embodies the type (Type1 & Type2) | Type4. Furthermore, if hasType4(data) also returns true, then within the subsequent if, data aligns with the type Type1 & Type2 & Type4.

A similar logic applies to the second instance of the doSomething function, where the type guards are invoked in reverse order.

Edit

For further clarification, here is a refined example:

type Type1 = { /* ... */ };
type Type2 = { /* ... */ };
type Type3 = { /* ... */ };
type Type4 = { /* ... */ };

type CombinatoryType = Type1 & Type2 & Type3;

function hasType2(data: CombinatoryType): data is CombinatoryType & Type2 {
 // Checking for properties of Type2 within 'data'
 return (data as Type2) /* ... */ !== undefined;
}

function hasType4(data: CombinatoryType & Type2): data is CombinatoryType & Type2 & Type4 {
 // Verifying properties of Type4 within 'data'
 return (data as Type4) /* ... */ !== undefined;
}

function doSomething(data: CombinatoryType) {
  if (hasType2(data)) {
      // 'data' now matches the criteria of (Type1 & Type2) | Type4

      if (hasType4(data)) {
          // 'data' now takes the form of Type1 & Type2 & Type4
      }
  }
}

In the aforementioned sample, the functions hasType2 and hasType4 ascertain whether data possesses attributes assigned to Type2 and Type4 respectively. Upon confirmation, the type of data is constricted to CombinatoryType & Type2 and

CombinatoryType & Type2 & Type4

correspondingly. This mechanism permits maintaining existing type information while incorporating new types.

Answer 2

If my interpretation is accurate based on the code you provided, it seems like you are aiming to establish type guards for both Type2 and Type4 within the CombinatoryType type. Am I correct in this assumption?

If that's the case, here is an illustration for you:

function hasType2(data: CombinatoryType): data is (Type1 & Type2) | Type4 {
   return 'dog' in data;
}

function hasType4(data: CombinatoryType): data is (Type1 & (Type2 | Type3)) & Type4 {
   return 'elephant' in data;
}

In the snippet above, we utilize the in operator to verify the existence of the dog property in data for hasType2, and similarly for the elephant property in data for hasType4.

The in operator acts as a type guard by confirming whether a property exists within an object.

The syntax data is (Type1 & Type2) | Type4 and data is (Type1 & (Type2 | Type3)) & Type4 serves to inform TypeScript that if the function returns true, then data conforms to the specified type.

These type guards can now be applied within your doSomething function:

function doSomething(data: CombinatoryType) {
   if (hasType2(data)) {
       // 'data' now falls under the type: (Type1 & Type2) | Type4

       if (hasType4(data)) {
           // 'data' now fits the mold of Type1 & Type2 & Type4
       }
   }
}

In this scenario, if hasType2(data) yields true, then within the initial if, data embodies the type (Type1 & Type2) | Type4. Furthermore, if hasType4(data) also returns true, then within the subsequent if, data aligns with the type Type1 & Type2 & Type4.

A similar logic applies to the second instance of the doSomething function, where the type guards are invoked in reverse order.

Edit

For further clarification, here is a refined example:

type Type1 = { /* ... */ };
type Type2 = { /* ... */ };
type Type3 = { /* ... */ };
type Type4 = { /* ... */ };

type CombinatoryType = Type1 & Type2 & Type3;

function hasType2(data: CombinatoryType): data is CombinatoryType & Type2 {
 // Checking for properties of Type2 within 'data'
 return (data as Type2) /* ... */ !== undefined;
}

function hasType4(data: CombinatoryType & Type2): data is CombinatoryType & Type2 & Type4 {
 // Verifying properties of Type4 within 'data'
 return (data as Type4) /* ... */ !== undefined;
}

function doSomething(data: CombinatoryType) {
  if (hasType2(data)) {
      // 'data' now matches the criteria of (Type1 & Type2) | Type4

      if (hasType4(data)) {
          // 'data' now takes the form of Type1 & Type2 & Type4
      }
  }
}

In the aforementioned sample, the functions hasType2 and hasType4 ascertain whether data possesses attributes assigned to Type2 and Type4 respectively. Upon confirmation, the type of data is constricted to CombinatoryType & Type2 and

CombinatoryType & Type2 & Type4

correspondingly. This mechanism permits maintaining existing type information while incorporating new types.

Is it possible to create a typeguard for a complex combinatory type while retaining the existing type information?

Answer №1

Edit

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