我基本上被要求获取字符串的Unicode值,将其乘以10%并添加对象当前具有的任何级别。这令人沮丧,因为事实证明我的逻辑已经包括代码但我仍然得到一个错误:预期:< 0>但是:< 8>。任何建议,也许这只是我必须在逻辑中做出的细微差别,虽然我很确定它是对的。 记下getLevel方法,因为错误在哪里
public class PouchCreature implements Battleable {
private String name;
private int strength;
private int levelUps;
private int victoriesSinceLevelUp;
/**
* Standard constructor. levelUps and victoriesSinceLevelUp start at 0.
*
* @param nameIn desired name for this PouchCreature
* @param strengthIn starting strength for this PouchCreature
*/
public PouchCreature(String nameIn, int strengthIn) {
this.name = nameIn;
this.strength = strengthIn;
this.levelUps = 0;
this.victoriesSinceLevelUp = 0;
}
/**
* Copy constructor.
*
* @param other reference to the existing object which is the basis of the new one
*/
public PouchCreature(PouchCreature other) {
this.name=other.name;
this.strength=other.strength;
this.levelUps=other.levelUps;
this.victoriesSinceLevelUp=other.victoriesSinceLevelUp;
}
/**
* Getter for skill level of the PouchCreature, which is based on the
* first character of its name and the number of levelUps it has.
* Specifically, the UNICODE value of the first character in its name
* taken %10 plus the levelUps.
*
* @return skill level of the PouchCreature
*/
public int getLevel() {
int value = (int)((int)(getName().charAt(0)) * 0.1);
return value + this.levelUps;
}
答案 0 :(得分:0)
您已经说过,您应该将该值增加10%。但是,你实际上做的是通过仅占用10%来减少 90%(然后将其截断为67.0 * 0.1 = 6.7
)。 int
,截断为6
时为0.1
。
将1.1
更改为int value = (int)((int)(getName().charAt(0)) * 1.1);
// --------------------------------------------^
至增加 10%:
getName()
如果"Centaur"
返回C
(例如),67
的Unicode值为value
,73
最终为{{1} }}
答案 1 :(得分:0)
我们需要查看您正在调用该类的代码并生成错误消息。为什么期待0? 8似乎是您提供的信息的有效返回值。