Combining types: unable to utilize the second optional type within a for loop

Question

Combining types: unable to utilize the second optional type within a for loop

I am facing an issue while looping through an array due to the union type. I am wondering what I have overlooked in the code below that is causing Visual Studio Code to not recognize the second optional type for that specific array.

class Menu {
  // name of the menu
 name: string;
  // list of the ingredients in this menu
  // or a list of submenu with its ingredients
 list: string[] | Menu[];
 hasSubList?: boolean;
}

...
menu: Menu[];

ngOnInit() {
  this.loadMenu();
}
ngAfterContentInit() {
  this.print();
}
loadMenu(): void {
  this.menu = [
   {
     name: "toast" 
     list: [
      "slide of bread",
     ],
   },
   {
     name: "Crumble eggs on taste",
     hasSubList: true;
     list: [
      {
        name: "Eggs",
        list: [
         {
           "Eggs",
           "pepper",
           "a pinch of salt",
          }
         ],
       },
       {
        name: "Toast",
        list: [
          "a slide of bread"
         ],
        },
       ],
   },
 ];
}

this print(): void {
  for(let i=0; i<this.menu.length;i++){
    let item = this.menu[i];
    console.log(item.name);
    for(let j=0; i<item.list.length; j++){
      let list = item.list[j];
      if(item.hasSubList) {
         // HERE
         // console intellsense says 
         // "property 'list' does not exist on type 'string | Menu'
         // "property 'list' does not exist on type 'string'
        for(let k=0; k< list.list.length; k++}(
          console.log(list.list[k]);
        }
      } else {
       console.log(list);
      }
}

Recapping the message from the intellsense:

"property 'list' does not exist on type 'string | Menu'
"property 'list' does not exist on type 'string'

Why did it fail to check for Menu? Because 'list' exists as type 'Menu'

angular typescript union-types

Answer 1

Answer №1

As pointed out by @jacobm, the compiler is unable to determine the actual type of the union type based solely on hasSubList.

In order for the compiler to understand, you must provide a way to discriminate it. One approach is to check the type of the item itself without relying on the hasSubList property.

For example:

for(let i=0; i<this.menu.length;i++){
    let item = this.menu[i];

    for(let j=0; i<item.list.length; j++){
        let list = item.list[j];

        // Using the typeof operator can help distinguish the union type
        if(typeof item !== 'string') {
            for(let k=0; k< list.list.length; k++}(
                console.log(list.list[k]);
            }
        } else {
            console.log(list);
        }
    }
}

Another alternative is to utilize for..of loops instead of traditional for loops. While this may be a matter of personal preference, using for..of loops can be more concise and improve readability when the loop index is not needed.

for (let item of menu) {
    for (let list of item.list) {
        if(item !== 'string') {
            for(let childItem of list.list) {
                console.log(childItem);
            }
        } else {
            console.log(list);
        }
    }
}

Answer 2

As pointed out by @jacobm, the compiler is unable to determine the actual type of the union type based solely on hasSubList.

In order for the compiler to understand, you must provide a way to discriminate it. One approach is to check the type of the item itself without relying on the hasSubList property.

For example:

for(let i=0; i<this.menu.length;i++){
    let item = this.menu[i];

    for(let j=0; i<item.list.length; j++){
        let list = item.list[j];

        // Using the typeof operator can help distinguish the union type
        if(typeof item !== 'string') {
            for(let k=0; k< list.list.length; k++}(
                console.log(list.list[k]);
            }
        } else {
            console.log(list);
        }
    }
}

Another alternative is to utilize for..of loops instead of traditional for loops. While this may be a matter of personal preference, using for..of loops can be more concise and improve readability when the loop index is not needed.

for (let item of menu) {
    for (let list of item.list) {
        if(item !== 'string') {
            for(let childItem of list.list) {
                console.log(childItem);
            }
        } else {
            console.log(list);
        }
    }
}

Answer 3

Answer №2

Given that list[j] is of type string | Menu, any code utilizing it must be able to handle both strings and Menus, not just one or the other. It seems like you are assuming that if hasSubList is true, then list will always contain a list of menus. If this is your intention, you can cast item to a menu using list as Menu. However, it is generally recommended to avoid such casts as they limit the ability of the type checker to catch potential errors in your code.

Answer 4

Given that list[j] is of type string | Menu, any code utilizing it must be able to handle both strings and Menus, not just one or the other. It seems like you are assuming that if hasSubList is true, then list will always contain a list of menus. If this is your intention, you can cast item to a menu using list as Menu. However, it is generally recommended to avoid such casts as they limit the ability of the type checker to catch potential errors in your code.

Answer 5

Answer №3

The error message occurs because Typescript is unable to determine the specific type you are using when you assign it as string | Menu[]. Using a type cast is one way to resolve this issue.

for (let j = 0; i < item.list.length; j++) {
    if (item.hasSubList) {
      let list = item.list[j] as Menu; // use 'as'     
      for (let k = 0; k < list.list.length; k++) {
        console.log(list[k]);
      }
    } else {
      let list = item.list[j];  
      console.log(list);
    }
  }

Answer 6

The error message occurs because Typescript is unable to determine the specific type you are using when you assign it as string | Menu[]. Using a type cast is one way to resolve this issue.

for (let j = 0; i < item.list.length; j++) {
    if (item.hasSubList) {
      let list = item.list[j] as Menu; // use 'as'     
      for (let k = 0; k < list.list.length; k++) {
        console.log(list[k]);
      }
    } else {
      let list = item.list[j];  
      console.log(list);
    }
  }

Combining types: unable to utilize the second optional type within a for loop

Answer №1

Answer №2

Answer №3

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