TypeScript claims no error even if parameter has the wrong type

Question

I have a function that takes a dictionary as the first argument. This dictionary has string as keys, and functions as values. The problem is that, if a function in the dictionary has a bad signature, TypeScript does not complain!

A piece of code is worth 1000 words. Here's my main.ts:

interface MyMap {
    [key: string]: (x: number) => void;
}

function f(map: MyMap, key: string, x: number) : void{
    map.hasOwnProperty(key) ? map[key](x) : console.log(x)
}

const stupidMap: MyMap = {
    'square': (x) => {
        console.log(x*x);

        return x*x; // This function should return void!!
    },
    'double': (x) => {
        console.log(x+x);
    }
}

f(stupidMap, 'square', 5) // Prints 25
f(stupidMap, 'double', 5) // Prints 10

I compile it with tsc main.ts, and I get no error whatsoever. tsc --version prints Version 3.7.2. I have two questions:

1. Why don't I get any error?
2. Am I missing some compile flags that would cause this error to show up?

Any insight would be very appreciated. Thanks!

Answer 1

没有问题，因为(x: number) => void可分配给(x: number) => number. 这是一个证明：

type F = (x: number) => void
type Z = (x: number) => number

type ZextendsF = Z extends F ? true : false // evaluate to true 

这个事实对于程序流程来说是完全没问题的。如果你的界面说 - 我需要一个不返回的函数，那么如果我传递一个返回某些东西的函数，它完全可以，因为我永远不会使用这个返回数据。它是类型安全的，无需担心。

有关函数可分配性的更多详细信息 -比较两个函数。还有更多关于 TypeScript 类型行为和关系类型可分配性的细节。