英文:
Generic intersection results in "'T' could be instantiated with an arbitrary type" error
问题
以下是您提供的代码的翻译部分：
通用条件语句的评估被延迟执行，因此我正在尝试使用交集来提示编译器我的期望类型：type ArgType<T> = T & (T extends MyType ? ({myArg?: MyInterface<T>}): never)然而，当我尝试调用一个访问上述参数类型的函数时，我遇到了以下错误：'T' 可以用任意类型实例化，这个类型可能与 '{ myArg: this; }' 不相关。完整代码：type MyType = 'foo';interface MyInterface<T extends MyType> {myField: T;}type ArgType<T> = T & (T extends MyType ? ({myArg?: MyInterface<T>}): never)function myFunc<T>(arg: ArgType<T>) {return arg;}class MyClass<T extends MyType> implements MyInterface<T> {constructor(readonly myField: T) {}invoke() {const result = myFunc({myArg: this});}}
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英文:
Generic conditionals have their evaluation deferred, so I am attempting to use an intersection to hint to the compiler my desired type:
type ArgType<T> = T & (T extends MyType ? ({myArg?: MyInterface<T>}): never)
However, when I try to invoke a function that access an argument typed to the above, I am getting this error:
> 'T' could be instantiated with an arbitrary type which could be unrelated to '{ myArg: this; }'
Full code:
type MyType = 'foo';interface MyInterface<T extends MyType> {myField: T;}type ArgType<T> = T & (T extends MyType ? ({myArg?: MyInterface<T>}): never)function myFunc<T>(arg: ArgType<T>) {return arg;}class MyClass<T extends MyType> implements MyInterface<T> {constructor(readonly myField: T) {}invoke() {const result = myFunc({myArg: this});}}
答案1
得分: 0
type MyType = 'foo';
interface MyInterface
myField: T;
}
type ArgType
myArg?: MyInterface
} : never;
function myFunc
return arg;
}
class MyClass
constructor(readonly myField: T) {}
invoke() {
const result = myFunc
myArg: this
});
}
}
英文:
type MyType = 'foo';interface MyInterface<T extends MyType> {myField: T;}type ArgType<T> = T extends MyType ? {myArg?: MyInterface<T>;} : never;function myFunc<T>(arg: ArgType<T>) {return arg;}class MyClass<T extends MyType> implements MyInterface<T> {constructor(readonly myField: T) {}invoke() {const result = myFunc<MyType>({myArg: this});}}
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