TypeScript does not enforce type check for function arguments

Question

I had an impression that TypeScript allowed you to take a valid JavaScript program and "enforce" types by making a few key symbols type-safe. This would propagate the types through all the usages, and make you update all the symbol references.

This seems to be incorrect. In the following example the makeDiv function is calling a typed make function without checking the parameter type.

// strongly typed arguments
function make(tagName: string, className: string) {
    alert ("Make: " + className.length);
}

// no typing
function makeDiv(className) {
    // calling the typed function without type checking
    return make("div", className);
}

makeDiv("test"); 
makeDiv(6); // bang!

Do I miss something here? Is there is a way to enforce "stricter" type checking? Or is this a design decision made by TypeScript creators?

basarat · Accepted Answer · 2013-08-19 04:23:58Z

5

This is a design decision. Anything not explicitly typed becomes implicitly typed to be of type any. any is compatible with all types.

var x:any = 123;
x = "bang";

To prevent implicit typing of a variable to be any there is a compiler flag (--noImplicitAny) starting with TypeScript 0.9.1

If you compile with this option your code will not compile unless you do:

// You need to explicitly mention when you want something to be of any type. 
function makeDiv(className:any) {
    // calling the typed function without type checking
    return make("div", className);
}

answered Aug 19, 2013 at 4:23

basarat

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2 Comments

Vitaly Brusentsev Over a year ago

Thanks @basarat. What I didn't understand was that any is compatible with all types (as you pointed out) in two ways. I mentally associated it with C# object, which could be downcasted from string, but not the other way around.

Dylanthepiguy Over a year ago

Actually, anything not explicitly typed has a type inferred if possible. E.g. var x = 3 becomes a number, and function a() { alert('abc') } returns void. For some reason you are allowed to return the void value that a() returns from another function, i.e. function b() { return a() } but b() is implied to return void

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