Operator cannot be applied to types typescript

Last updated: Jan 23, 2023
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# Operator ‘+’ cannot be applied to types ‘Number’ and number

The error «Operator ‘+’ cannot be applied to types ‘Number’ and ‘number'» occurs when we use the Number type instead of number (lowercase n ).

To solve the error, make sure to use the number type with lowercase n in your TypeScript code.

Here is an example of how the error occurs.

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const num: Number = 100; // ⛔️ Error: Operator '+' cannot be applied // to types 'Number' and 'number'.ts(2365) console.log(num + 100);

We used the Number , non-primitive object type to type the num variable, which caused the error.

We can’t use the addition operator (+) to add the non-primitive object type to a value of type number .

# Always use the number type when typing numbers in TypeScript

To solve this, always use the number type when typing numbers in TypeScript.

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const num: number = 100; console.log(num + 100); // 👉️ 200

The type in the example can be inferred based on the provided value, so we don’t even have to set it.

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// 👇️ const num: 100 const num = 100; console.log(num + 100); // 👉️ 200

We used the primitive number type which resolved the error.

# Convert the object type to a primitive

If you don’t have access to the code that uses the Number object type, convert the object type to a primitive number by passing it to the Number() function.

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const num = 100 as Number; console.log(Number(num) + 100); // 👉️ 200

You might also see examples online that use the unary plus (+) operator to convert a value to a number .

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const num = 100 as Number; console.log(+num + 100);

The error was caused because there is a difference between the primitive number , string and boolean types and the non-primitive Number , String , Boolean , Object , etc.

The non-primitive types are objects and should never be used when typing values in TypeScript.

The TypeScript best practices documentation warns to never use the Number , String , Boolean , Symbol or Object non-primitive objects when typing values in your TypeScript code.

Instead, you should be using number , string , boolean , symbol and Record types.

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

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Operator ‘+’ cannot be applied to types ‘T’ and ‘T’ #41449

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I think @Jameskmonger is right that this is a compiler bug. The type T is definitely known and constrained by the compiler to be a specific type, that’s important to how generics fundamentally work and why the compiler refuses to compile even if you force it to accept the addition, e.g.:

type Numeric = bigint | number; function sum(a: T, b: T) < return a + b; > console.log(sum(1, 2)) console.log(sum(1n, 2n)) console.log(sum(1n, 2))

Note that the last sum is refused by the compiler (not at runtime).

@Jameskmonger if T extends string | number then T can be string | number which means one value of type T can be string a different value of type T can be a number .

It really sounds like you should use overloads here rather than a type parameter.

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Operator ‘+’ cannot be applied to types ‘T’ and ‘T’. #12410

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TypeScript Version: 2.1.1 / nightly (2.2.0-dev.201xxxxx)

function addT extends string | number>(x: T, y: T): T  return x + y; > addstring>("a", "b"); addnumber>(5, 3); add("a", "b"); add(5, 3);

Expected behavior:

Code to compile fine, because T is of type string | number and therefore can have the + operator applied.

Actual behavior:

Operator ‘+’ cannot be applied to types ‘T’ and ‘T’.

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error TS2365: Operator ‘+’ cannot be applied to types ‘Number’ and ‘Number’. #2031

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var a: Number = new Number(1); var b: Number = new Number(2); var c: number = 3; console.log(a+b); console.log(b+c);

When compiled, the compiler produces these errors:

index.ts(4,13): error TS2365: Operator '+' cannot be applied to types 'Number' and 'Number'. index.ts(5,13): error TS2365: Operator '+' cannot be applied to types 'Number' and 'number'.

But if I don’t use —noEmitOnError, the compiler produces this index.js:

var a = new Number(1); var b = new Number(2); var c = 3; console.log(a + b); console.log(b + c);

Which I can run with node and get expected results:

Why does the compiler refuse to allow addition of two Numbers, or a Number and a number?

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danquirk added the By Design Deprecated — use «Working as Intended» or «Design Limitation» instead label Feb 13, 2015

@danquirk, does this seem correct to you?

«The binary + operator requires both operands to be of the Number primitive type or. «
index.ts(4,13): error TS2365: Operator ‘+’ cannot be applied to types ‘Number’ and ‘Number’.

The language of section 4.15.2 is confusing because it only says Number (with a capital N) but then says primitive type , which implies number (with a lowercase n). If the Typescript team really is going to insist that only number types can be added, and that Number types cannot, then I think that section of the spec should be clarified.

But I want to protest this stance.

First, you can’t say ‘in practice people aren’t creating number via new Number’, because you obviously don’t have full knowledge of all javascript practice. I am a user (and minor contributor) to node-java, a bridge API that allows node.js applications to launch an embedded JVM and execute Java code. The node-java API returns javascript Number objects everywhere a Java method returns a numeric type.

Second, yes, you provided an artificial example how someone can create a var with a Number interface, but do you think that people do that in practice? Do you think that a programmer who did that would complain to you that applying operator+ to his var yielded incorrect results, and deny that his code was nonsensical?

Finally, I want typescript to catch at compile-time errors in use of types that are likely to result in errors at runtime, and not yield compile-time errors when the emitted code is unlikely to lead to a compile time error. I know there is a probably a lot of gray area here, but when in doubt, typescript should do no harm, i.e not prevent a programmer from using correct Javascript. If the Javascript spec disallowed adding two Numbers, or adding a Number to a number, then Typescript clearly should too. But it is wrong for Typescript to disallow something so simple as adding two numbers.

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Operator cannot be applied to types typescript

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Error «Operator ‘===’ cannot be applied to types» with enum type #11533

Error «Operator ‘===’ cannot be applied to types» with enum type #11533

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