如果以前曾问过这个问题,我很抱歉,但我真的不知道要搜索什么。
无论如何,我正在制作一个数学包,许多类扩展了Function:
package CustomMath;
@SuppressWarnings("rawtypes")
public abstract class Function <T extends Function> {
public abstract Function getDerivative();
public abstract String toString();
public abstract Function simplify();
public abstract boolean equals(T comparison);
}
我想比较函数以确定它们是否相等。如果它们来自同一个类,我想使用它的特定比较方法,但如果它们是不同的类,我想返回false。这是我目前的课程之一:
package CustomMath;
public class Product extends Function <Product> {
public Function multiplicand1;
public Function multiplicand2;
public Product(Function multiplicand1, Function multiplicand2)
{
this.multiplicand1 = multiplicand1;
this.multiplicand2 = multiplicand2;
}
public Function getDerivative() {
return new Sum(new Product(multiplicand1, multiplicand2.getDerivative()), new Product(multiplicand2, multiplicand1.getDerivative()));
}
public String toString() {
if(multiplicand1.equals(new RationalLong(-1, 1)))
return String.format("-(%s)", multiplicand2.toString());
return String.format("(%s)*(%s)", multiplicand1.toString(), multiplicand2.toString());
}
public Function simplify() {
multiplicand1 = multiplicand1.simplify();
multiplicand2 = multiplicand2.simplify();
if(multiplicand1.equals(new One()))
return multiplicand2;
if(multiplicand2.equals(new One()))
return multiplicand1;
if(multiplicand1.equals(new Zero()) || multiplicand2.equals(new Zero()))
return new Zero();
if(multiplicand2.equals(new RationalLong(-1, 1))) //if one of the multiplicands is -1, make it first, so that we can print "-" instead of "-1"
{
if(!multiplicand1.equals(new RationalLong(-1, 1))) // if they're both -1, don't bother switching
{
Function temp = multiplicand1;
multiplicand1 = multiplicand2;
multiplicand2 = temp;
}
}
return this;
}
public boolean equals(Product comparison) {
if((multiplicand1.equals(comparison.multiplicand1) && multiplicand2.equals(comparison.multiplicand2)) ||
(multiplicand1.equals(comparison.multiplicand2) && multiplicand2.equals(comparison.multiplicand1)))
return true;
return false;
}
}
我该怎么做?
答案 0 :(得分:1)
使用泛型,您可以保证equals方法仅适用于类型'T',在本例中为'Product'。你不能传递另一种类型。
另一种可能性是在classe函数定义:
public abstract boolean equals(Function comparison);
在classe Product中,对象比较为comparison instanceof Product
答案 1 :(得分:1)
覆盖Object.equals(Object)方法。您不需要在这里使用泛型。它的身体看起来像这样
if (other instanceof Product) {
Product product = (Product) other;
// Do your magic here
}
return false;