No error report when using compound assignment operator with literal types.

[Veetaha]

TypeScript Version: 3.4.2

• literal type shorthand assignment
• shorthand assignment

Code

let ten = 10 as const;

// no error reports
ten += 2;
ten -= 2;
ten *= 2;
ten /= 2;
ten **= 2;
ten %= 2;
ten |= 2;
ten &= 2;
ten ^= 2;
ten >>= 2;
ten >>>= 2;
ten <<= 2;

// all errors as expected
ten = ten + 2;
ten = ten - 2;
ten = ten * 2;
ten = ten / 2;
ten = ten ** 2;
ten = ten % 2;
ten = ten | 2;
ten = ten & 2;
ten = ten ^ 2;
ten = ten >> 2;
ten = ten >>> 2;
ten = ten << 2;

let pinkie = 'Pinkie' as const;

// no error report
pinkie += 'Pie';

// error as expected
pinkie = pinkie + 'Pie';

Expected behavior:

Shorthand assignment to a literal type where the resulting assigned value may not always be of that literal type should generate a compile-time error likewise its expanded form.

Actual behavior:
There is no error report for using shorthand assignment operators that might store an invalid value for that literal type. But the expanded form of those operators works as expected.

Playground Link:

Klick me

Related Issues:

[Veetaha]

Quick inline proposal: compute literal types expressions at compile-time so that ten * 2 would be of type 20 but not number or type of 'Pinkie' + 'Pie' expression would be 'PinkiePie' instead of string.

[DanielRosenwasser]

Potentially related to #13865, but I feel like we have another issue about this.

[RyanCavanaugh]

Duplicate #14745

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.