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Question

How can I make my typescript compiler happy without changing the interface and typeof argument I'm receiving in function test.

Error in function test:-

"Property 'method2' does not exist on type 'xyz'. Did you mean 'method1'?"

 interface xyz { method1(): string; } class abc implements xyz { method1() { return "abc"; } method2() { return "new method"; } } function test(arg: xyz) { alert(arg.method2()); } 

Link

Answer 1

You can use a type guard to change the type that is seen at the compiler when you want to access the other fields:

function isAbc(arg: xyz | abc): arg is abc {
    return (<abc>arg).method2 !== undefined;
}

function test(arg: xyz) {
    if (isAbc(arg)) {
        // here the type of `arg` is `abc`
        alert(arg.method2());
    }
}

Answer 2

Actually you can't.

Why ?

To make your code to pass compiler you need either add the method2 into the interface xyz or change the type parameter to accept the type abc . But you don't want neither.

Answer 3

After going through documents, got to know about Type assertions, which helped me to compile that small piece of code successfully.

function test(arg: xyz) {
   var arg2 = <abc>arg; 
   alert(arg2.method2());
}

https://www.typescriptlang.org/docs/handbook/basic-types.html