Java 加法多项式

我需要一些帮助。我正在做测试驱动开发。以下是测试：

@Test public void add01() throws TError { assertEquals(new Term(10,5),  new Term(4,5).add(new Term(6,5))); }
@Test public void add02() throws TError { assertEquals(new Term(6,5),   new Term(0,5).add(new Term(6,5))); }
@Test public void add03() throws TError { assertEquals(new Term(2,5),   new Term(-4,5).add(new Term(6,5))); }
@Test public void add04() throws TError { assertEquals(new Term(10,0),  new Term(4,0).add(new Term(6,0))); }
@Test public void add05() throws TError { assertEquals(new Term(4,5),   new Term(4,5).add(new Term(0,5))); }
@Test public void add06() throws TError { assertEquals(new Term(-2,5),  new Term(4,5).add(new Term(-6,5))); }

@Test (expected = IncompatibleTerms.class)
public void add07() throws TError { t = new Term(4,5).add(new Term(6,0)); }

@Test (expected = CoefficientOverflow.class)
public void add08() throws TError { t = new Term(min,4).add(new Term(-1,4)); }

@Test public void add09() throws TError { assertEquals(new Term(min,4), new Term(min+1,4).add(new Term(-1,4))); }
@Test public void add10() throws TError { assertEquals(new Term(-11,4), new Term(-10,4).add(new Term(-1,4))); }
@Test public void add11() throws TError { assertEquals(new Term(-2,4),  new Term(-1,4).add(new Term(-1,4))); }
@Test public void add12() throws TError { assertEquals(new Term(-1,4),  new Term(0,4).add(new Term(-1,4))); }
@Test public void add13() throws TError { assertEquals(Term.Zero,       new Term(1,4).add(new Term(-1,4))); }
@Test public void add14() throws TError { assertEquals(new Term(9,4),   new Term(10,4).add(new Term(-1,4))); }
@Test public void add15() throws TError { assertEquals(new Term(max,4), new Term(max-1,4).add(new Term(1,4))); }

@Test (expected = CoefficientOverflow.class)
public void add16() throws TError { t = new Term(max,4).add(new Term(1,4)); }

//Using domain-specific knowledge:  addition should be commutative

@Test public void add17() throws TError { assertEquals(new Term(-1,2).add(new Term(1,2)),       new Term(1,2).add(new Term(-1,2))); }
@Test public void add18() throws TError { assertEquals(new Term(min+1,2).add(new Term(-1,2)),   new Term(-1,2).add(new Term(min+1,2))); }
@Test public void add19() throws TError { assertEquals(new Term(min,2).add(Term.Zero),      Term.Zero.add(new Term(min,2))); }
@Test public void add20() throws TError { assertEquals(new Term(min,0).add(Term.Unit),      Term.Unit.add(new Term(min,0))); }
@Test public void add21() throws TError { assertEquals(new Term(max,0).add(Term.Zero),      Term.Zero.add(new Term(max,0))); }
@Test public void add22() throws TError { assertEquals(new Term(max-1,2).add(new Term(1,2)),    new Term(1,2).add(new Term(max-1,2))); }
@Test public void add23() throws TError { assertEquals(new Term(max,2).add(new Term(min,2)),    new Term(min,2).add(new Term(max,2))); }

这是我到目前为止在“Andrej Gajduk”的帮助下添加的方法

这些测试失败=2,5,8,12,16,19,21。但我对第17-23次考试有疑问；我知道他们中的一些人通过了，但我认为他们不应该通过，因为这就像x+y=y+x，但我不知道如何实现它

我需要一些帮助/指导来解释为什么测试失败

对于测试8和16，我尝试了下面的代码。它完成了任务，但在其他测试中会产生更多错误。所以所有的失败加起来变成2，5，8，12，16，19，21+20，23（两个额外的失败）

---

此链接显示检测溢出的另一种方法：

我认为这是因为当coef为0时，无论是这个还是整个语句都假设为零，所以你需要检查这个，并使用零的替代方案

if(this.coef ==0)
{//return Term t = new Term(that.coef, that.expo);}

else
{//return Term t = new Term(this.coef, this.expo);}
}

---

我希望这有帮助

您的

equals

方法不正确。请更新如下：

@Override
public boolean equals(Object that) {
    if (that == null) {
        return false;
    }else if (this.getClass() != that.getClass()) {
        return false;
    }else if (this == that) {
        return true;
    }else{
        Term term = (Term) that;
        if ((term.coef) == (this.coef) && (term.expo) == (this.expo)) {
            return true;
        }
    }
    return false;
}

public Term add(Term that) throws IncompatibleTerms, CoefficientOverflow{
    //handle addition with term zero
        if(this.equals(Term.Zero)){
        return that;
    }else if(Term.Zero.equals(that)){
        return this;
    }

    if (this.expo != that.expo) throw new IncompatibleTerms();
    if((this.coef>= 0 && that.coef >= 0 && this.coef + that.coef <0) ||
            (this.coef<= 0 && that.coef <= 0 && this.coef + that.coef >0)){
                         throw new CoefficientOverflow();
        }    
    return new Term (this.coef + that.coef,expo);
}

要使大多数测试用例通过，请更新

add

方法，如下所示：

@Override
public boolean equals(Object that) {
    if (that == null) {
        return false;
    }else if (this.getClass() != that.getClass()) {
        return false;
    }else if (this == that) {
        return true;
    }else{
        Term term = (Term) that;
        if ((term.coef) == (this.coef) && (term.expo) == (this.expo)) {
            return true;
        }
    }
    return false;
}

public Term add(Term that) throws IncompatibleTerms, CoefficientOverflow{
    //handle addition with term zero
        if(this.equals(Term.Zero)){
        return that;
    }else if(Term.Zero.equals(that)){
        return this;
    }

    if (this.expo != that.expo) throw new IncompatibleTerms();
    if((this.coef>= 0 && that.coef >= 0 && this.coef + that.coef <0) ||
            (this.coef<= 0 && that.coef <= 0 && this.coef + that.coef >0)){
                         throw new CoefficientOverflow();
        }    
    return new Term (this.coef + that.coef,expo);
}

public Term add（抛出不兼容rms、系数overflow的术语）{
//处理零项加法
如果（此项等于（项零））{
归还；
}else如果（项0等于（that））{
归还这个；
}
如果（this.expo！=that.expo）抛出新的不兼容rms（）；
如果（（this.coef>=0&&that.coef>=0&&this.coef+that.coef），您可以发布您的
术语.equals（）
method？这不是测试驱动开发。这不是测试驱动开发吗？@Lax：我在答案中做了一些更新。如果这没有帮助，请检查并让我知道。感谢您的帮助，它确实通过了大多数测试，除了7、8和16。我试图找出y 7失败，因为不同指数的条件是correct@Lax：对于me、 所有的测试用例都通过了。你有什么特殊的构造函数方法吗？公共项（int c，int e）抛出NegativeExponent{if（e<0）抛出新的NegativeExponent（）；coef=c；expo=（coef==0）？1:e；}@Lax：我使用了相同的方法，所有测试用例都通过了。只是想检查一下，您是否使用了答案中更新的
equals
方法。有一个问题，我在某个时候更新了它。