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No error report when using compound assignment operator with literal types. #30783
Description
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:
Related Issues:
Activity
Veetaha commentedon Apr 5, 2019
Quick inline proposal: compute literal types expressions at compile-time so that
ten * 2would be of type20but notnumberor type of'Pinkie' + 'Pie'expression would be'PinkiePie'instead ofstring.[-]No error report when using shorthand assignment operator with literal types.[/-][+]No error report when using compound assignment operator with literal types.[/+]DanielRosenwasser commentedon Apr 6, 2019
Potentially related to #13865, but I feel like we have another issue about this.
RyanCavanaugh commentedon Apr 9, 2019
Duplicate #14745
typescript-bot commentedon Apr 12, 2019
This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.