我想实现这样的目标:
class TestClass {
someMethod(stringParameter: string): void {
alert("Variant #1: stringParameter = " + stringParameter);
}
someMethod(numberParameter: number, stringParameter: string): void {
alert("Variant #2: numberParameter = " + numberParameter + ", stringParameter = " + stringParameter);
}
}
var testClass = new TestClass();
testClass.someMethod("string for v#1");
testClass.someMethod(12345, "string for v#2");
这是一个我不想做的例子(我真的很讨厌在 JS 中重载 hack 的那部分):
class TestClass {
private someMethod_Overload_string(stringParameter: string): void {
// A lot of code could be here... I don't want to mix it with switch or if statement in general function
alert("Variant #1: stringParameter = " + stringParameter);
}
private someMethod_Overload_number_string(numberParameter: number, stringParameter: string): void {
alert("Variant #2: numberParameter = " + numberParameter + ", stringParameter = " + stringParameter);
}
private someMethod_Overload_string_number(stringParameter: string, numberParameter: number): void {
alert("Variant #3: stringParameter = " + stringParameter + ", numberParameter = " + numberParameter);
}
public someMethod(stringParameter: string): void;
public someMethod(numberParameter: number, stringParameter: string): void;
public someMethod(stringParameter: string, numberParameter: number): void;
public someMethod(): void {
switch (arguments.length) {
case 1:
if(typeof arguments[0] == "string") {
this.someMethod_Overload_string(arguments[0]);
return;
}
return; // Unreachable area for this case, unnecessary return statement
case 2:
if ((typeof arguments[0] == "number") &&
(typeof arguments[1] == "string")) {
this.someMethod_Overload_number_string(arguments[0], arguments[1]);
}
else if ((typeof arguments[0] == "string") &&
(typeof arguments[1] == "number")) {
this.someMethod_Overload_string_number(arguments[0], arguments[1]);
}
return; // Unreachable area for this case, unnecessary return statement
}
}
}
var testClass = new TestClass();
testClass.someMethod("string for v#1");
testClass.someMethod(12345, "string for v#2");
testClass.someMethod("string for v#3", 54321);
TypeScript语言中如何实现方法重载?
最佳答案
根据规范,TypeScript 确实支持方法重载,但它非常笨拙,并且包含大量手动检查参数类型的工作。我认为这主要是因为在纯 JavaScript 中最接近方法重载的方法包括检查,并且 TypeScript 尽量不修改实际的方法体以避免任何不必要的运行时性能成本。
如果我理解正确,您必须首先为每个重载编写一个方法声明,然后一个方法实现检查其参数以确定调用哪个重载。实现的签名必须与所有重载兼容。
class TestClass {
someMethod(stringParameter: string): void;
someMethod(numberParameter: number, stringParameter: string): void;
someMethod(stringOrNumberParameter: any, stringParameter?: string): void {
if (stringOrNumberParameter && typeof stringOrNumberParameter == "number")
alert("Variant #2: numberParameter = " + stringOrNumberParameter + ", stringParameter = " + stringParameter);
else
alert("Variant #1: stringParameter = " + stringOrNumberParameter);
}
}
关于javascript - 如何在 TypeScript 中进行方法重载?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12688275/