打字稿：联合类型和三元运算符的类型推断

Question

The following typescript code compiles just fine:

let x: 0 | 1 = 0;

console.log(x);

const r = [true, false];
for (const y of r) {
    if (y || (x !== 1)) {
        x = 1;
    } else {
        x = 0;
    }

    console.log(x);
}

However this code, which is semantically equivalent does not:

let x: 0 | 1 = 0;

console.log(x);

const r = [true, false];
for (const y of r) {
    x = ((y || (x !== 1)) ? 1 : 0);
    console.log(x);
}

Error for the x !== 1 is:

7:17 error TS2367: This condition will always return 'true' since the types '0' and '1' have no overlap.

In both cases, running the compiled result yields the expected output showing that x does in fact take the value 0 and 1:

0
1
0

I understand the error stems from the fact that the compiler narrowed down the type 0 | 1 0 | 1 to just 0 in the second case. However, just by looking at the code it is clear that it could be that x is assigned 1 (even without looking at the condition). Thus, I would expect type inference to assume the most general type unless explicitly told otherwise (as it does in the first example). In fact, in line 1 I explicitly tell the compiler that I want the more general type: let x: 0 | 1 let x: 0 | 1 .

So my question would be if there is a sensible reason why the type inference behaves differently in case of the ternary operator?

Answer 1

The error is actually because the compiler has narrowed the type of x to 0 .

I'm not sure why you don't get the error in the other case - as I would expect it to be consistent in both cases - but I'm not smart enough to write the tools that do this stuff.

Example where the type of x is forced to be 0 | 1 0 | 1 rather than the compiler narrowing it to zero:

let x = 0 as 0 | 1;

console.log(x);

const r = [true, false];
for (const y of r) {
    x = ((y || (x !== 1)) ? 1 : 0);
    console.log(x);
}