Why is the function not being called when declaring a variable, resulting in a type error?

const fn1 = (arg1: {
    key: number,
})=>{
    console.log(arg1)
}

fn1({
 key: 1
})

const data = {
    key: 1,
    a: 1,
}
fn1(data)

fn1({
    key: 1,
    a: 1,
})
  • more info
  • Assistance needed: Why is the TypeScript function not called when declaring a variable, resulting in a type error?

Answer №1

function fn1(input: {
    key: number,
})=>{
    console.log(input)
}

When defining the parameter in this function, it specifies that it must have a key named "key" with a number value and nothing else. If you want to create a type for a parameter that can accept any key with a number value pair, you should update it as follows:

const fn1 = (input: {
    [key: string]: number
})=>{
    console.log(input)
}

Answer №2

Resolving the Issue

Upon examining the error message, it appears that:

The object literal can only define known properties, and 'a' is not present in the type '{ key: number; }'

This discrepancy arises between an object literal and a specifically named object.

Nevertheless, you have the option to modify your function to accommodate this change for successful compilation.

const func1 = <T extends {key: number}>(input: T)=>{
    console.log(input)
}

Revision

You may be wondering why this alteration is necessary, rather than how to implement it. This adjustment was actually introduced as a feature in TypeScript 1.6 to detect typographical errors made by developers in property names.

If preferred, you can enable the suppressImplicitAnyIndexErrors setting in your tsconfig file to ignore this issue. However, it is generally discouraged by TypeScript.

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