Should infer not undefined in "else" when checked in compound boolean expression in "if"

[yftachman]

🔎 Search Terms

"infer", "undefined", "compound", "boolean expression"

🕗 Version & Regression Information

i dont know if this is changed behaviour

⏯ Playground Link

https://www.typescriptlang.org/play?#code/JYOwLgpgTgZghgYwgAgCoQM5mQbwFDKHJgBcyWUoA5sgD7ICuIAJhDKBMwNx4C+eeGEwRhgAexDFMYAGLCAFHDLosAGmQAjMhrFiANhDggAlPiLIA9BeQAFKGI0GAtsgRwMEAkWAxk8jcgAZIHIAIRwAHRgxrhe5oQIEhj6EBF6YlTyAOQ+WcZxyLzIEHoesfFEVsgAkgByAMIA8gBKzQCi9ajIALTIIGLYTKzsIJzkABZiDHrMmiigMNBQnAXmiSBYmsgAvMiRYBGgrAAejTDZAIx5POb8AuZVAOpiUADW1K7unuY+fgHBeyiMTMFUs1iarQ6XV6-UGLDYHFmCyWK1BriS2ACu32h3hp3OWSuxhuRDuD2szzeHzcHgKv3k4SB5VBVgKRRKZRBFSqEPanR6fQGjHhIzGyKgy2YqyI602WMBByOEHxl2ubL4QA

💻 Code

interface Test {
    t: string | undefined;
}

function testFunc(a: Test, b: boolean){
    // Problem case
    if (b && !a.t) {
        console.log('if')
    } else {
        // INCORRECT - not undefined should be inferred
        const b = a.t.indexOf('1');
    }


    // Working case
    if (b && a.t) {
        // CORRECT - not undefined inferred
        const b = a.t.indexOf('1');
    }

    // Working case
    if (!a.t) {
        //
    } else {
        // CORRECT - not undefined inferred
        const b = a.t.indexOf('1');
    }
}

🙁 Actual behavior

a.t was checked to be undefined by if, but in else it is not inferred.

in test case 1, we can see that doing the reverse (checking for defined and checking in "if" for defined works correctly)
in test case 2, we can see that if its not a compound boolean expression it works correctly

🙂 Expected behavior

inferred type in else should be defined because it was checked in the if

[MartinJohns]

    // INCORRECT - not undefined should be inferred
   const b = a.t.indexOf('1');

If your boolean b is false, then you have no clue whether a.t is undefined or not. Why shouldn't it be undefined here?

This is a logic error on your side.

[mwlik]

That is precisely what Martin just said.

Here is a truth table to understand more the logic:

a.t	b	b && !a.t	Result in "if" block
1	1	false	Not executed
1	0	false	Not executed
0	1	true	Executed
0	0	false	Not executed

The last case is the case that the compiler takes into consideration.

[RyanCavanaugh]

The legal call

testFunc({ t: undefined }, false)

crashes this function.