key: string]: any; }' is not assignable to type 'T'.
'{ [key: string]: any; }' is assignable to the constraintof type 'T', but 'T' could be instantiated with a different subtype of constraint '{ [key: string]:
不知何故,我构建方法的方法有以下编译错误。Type '{ name: string; }' is not assignable to type 'A'. '{ name: string; }' is assignable to the constraintof type 'A', but 'A' could be instantiated with a different subtype of constraint 'Base
</ApolloConsumer>;}Type 'R & { apolloClient: ApolloClient<object>; }' is not'R & { apolloClient: ApolloClient<object>; }' is assignable to the constraint of type 'P', but 'P' couldbe i
string): T => ({});'{ value: string; }' is assignable to the constraintof type 'T', but 'T' could be instantiated with a different subtype of constraint '{ value: string;
但是在处理泛型类型时遇到了一些问题。 我有一个抽象的BaseModel,它有一个返回自身和子类型的create方法。为此,我说它将返回BaseModel。在具有从BaseModel扩展的泛型类型的存储库中,我包装了我的模型方法,并使用泛型类型作为返回值。但当我这样做时,它会给我以下的误差。'BaseModel' is assignable to the constrain
return fixtureStore[entityName](); // ERROR: Type 'OrderFixture' is not assignable to type 'T'.它会产生以下类型错误with a different subtype of constraint 'OrderFixture | DecisionFixture'.subtype of constraint 'OrderFixture | DecisionFi
有一个类,我想为它定义一个实例。看起来是这样的: funcOne:: (Real b) => a b -> a b我不能将(Real b)更改为(Floating b),因为其他实例也应该使用整数类型。但MyValue只适用于Floating。如果不是,那么如何才能将结果x*pi转换回与x参数相同的Real?如果类型是整型,那么发生什么并不重要,因
我不太确定错误消息是什么意思,但从逻辑上看,签名是完全有效的。这就是TS不支持的吗?'boolean' is assignable to the constraint of type 'isOutputOrdered_T', but 'isOutputOrdered_T' couldbe instantiated with a different subtype of constraint 'boolean'.ts(2322)