How can you pass the arguments of a function into another function without needing to assert the parameters?
Example:
function f(a:number, b:number){
let args:Parameters<typeof f> = arguments // Error: Type 'IArguments' is not assignable to type '[a: number, b: number]'.
}
Example:
function fx(a:number, b:number){
fy(...arguments); // Error: Type 'IArguments' is not assignable to type '[a: number, b: number]'.
}
function fy(a:number, b:number){
console.log(arguments);
}
At this time, TypeScript does not have a native way to achieve this functionality (without resorting to something like a type assertion, which is not recommended).
Currently, the arguments object is assigned the IArguments type, which resembles an array with its numeric index signature and a numeric length property. However, it lacks enough typing information to determine the argument types at specific indexes or the exact literal type of the length property:
function f(a: number, b: number) {
const arg0 = arguments[0];
// const arg0: any
const len = arguments.length;
// const len: number
}
In essence, while IArguments is similar to an array, it lacks the specificity of being tuple-like. There has been ongoing discussion on enhancing the typing of arguments in the issue tracker at microsoft/TypeScript#29055. It's possible that a specialized IArguments<T> type could be introduced in the future, functioning like a stripped-down tuple type without array-specific methods. This poses some challenges due to the current spread behavior only applicable to actual tuples as function arguments. So, creating a custom IArguments<T> may not seamlessly integrate with existing syntax, such as calling
f(...arguments)</code while preventing invalid operations like <code>arguments.map(x => x)As of now, this feature is not integrated into the language. To work around this limitation, one would need to rely on type assertions or similar techniques.
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