Invalid transform of async arrow expression containing super #46828
Description
Bug Report
🔎 Search Terms
super property setter getter method lowered async arrow function expression generator
🕗 Version & Regression Information
- This is the behavior in every version I tried
⏯ Playground Link
Playground link with relevant code
💻 Code
class Base {
set setter(x: string) {}
get getter() { return '' }
method(x: string) {}
static set setter(x: number) {}
static get getter() { return 0 }
static method(x: number) {}
}
class Derived extends Base {
a() { return async () => super.method('') }
b() { return async () => super.getter }
c() { return async () => super.setter = '' }
static a() { return async () => super.method(0) }
static b() { return async () => super.getter }
static c() { return async () => super.setter = 0 }
}🙁 Actual behavior
Here's the above code compiled by TypeScript with a target of ES2015:
class Base {
set setter(x) { }
get getter() { return ''; }
method(x) { }
static set setter(x) { }
static get getter() { return 0; }
static method(x) { }
}
class Derived extends Base {
a() { return () => __awaiter(this, void 0, void 0, function* () { return super.method(''); }); }
b() { return () => __awaiter(this, void 0, void 0, function* () { return super.getter; }); }
c() { return () => __awaiter(this, void 0, void 0, function* () { return super.setter = ''; }); }
static a() { return () => __awaiter(this, void 0, void 0, function* () { return super.method(0); }); }
static b() { return () => __awaiter(this, void 0, void 0, function* () { return super.getter; }); }
static c() { return () => __awaiter(this, void 0, void 0, function* () { return super.setter = 0; }); }
}The resulting code contains syntax errors:
Uncaught SyntaxError:
a() { return () => __awaiter(this, void 0, void 0, function* () { return super.method(''); }); }
^
'super' keyword outside a method
🙂 Expected behavior
I expected the uses of super inside async arrow expressions to be transformed into a helper function similarly to how TypeScript currently transforms super inside async methods. Perhaps TypeScript's output would look something like this instead:
class Base {
set setter(x) { }
get getter() { return ''; }
method(x) { }
static set setter(x) { }
static get getter() { return 0; }
static method(x) { }
}
class Derived extends Base {
a() {
const _super = Object.create(null, { method: { get: () => super.method } });
return () => __awaiter(this, void 0, void 0, function* () { return _super.method.call(this, ''); });
}
b() {
const _super = Object.create(null, { getter: { get: () => super.getter } });
return () => __awaiter(this, void 0, void 0, function* () { return _super.getter; });
}
c() {
const _super = Object.create(null, { setter: { get: () => super.setter, set: v => super.setter = v } });
return () => __awaiter(this, void 0, void 0, function* () { return _super.setter = ''; });
}
static a() {
const _super = Object.create(null, { method: { get: () => super.method } });
return () => __awaiter(this, void 0, void 0, function* () { return _super.method.call(this, 0); });
}
static b() {
const _super = Object.create(null, { getter: { get: () => super.getter } });
return () => __awaiter(this, void 0, void 0, function* () { return _super.getter; });
}
static c() {
const _super = Object.create(null, { setter: { get: () => super.setter, set: v => super.setter = v } });
return () => __awaiter(this, void 0, void 0, function* () { return _super.setter = 0; });
}
}