Java：toString 错误“错误：类型不兼容：int 无法转换为字符串”
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Java: toString error "error: incompatible types: int cannot be converted to String"
提问by Paincakes
I just need a quick help with my toString. This involves playing cards.
我只需要快速帮助我的toString. 这涉及打牌。
public class Card
{
//Data
private int rank;
private char rank2;
private char suit;
//more codes (constructor etc.)
//toString
public String toString ( )
{
if (rank == 11)
rank2 = 'J';
else if (rank == 12)
rank2 = 'Q';
else if (rank == 13)
rank2 = 'K';
else if (rank == 14)
rank2 = 'A';
else
return rank2 + suit; // <----error here.
}
I am trying to figure out how to return a rank and a suit. For example, if someone were to input 11 for the rank and 'S' for the suit, then I should return a JS. However, I am running into an error:
我想弄清楚如何归还军衔和西装。例如，如果有人要为等级输入 11，为西装输入“S”，那么我应该返回一个 JS。但是，我遇到了一个错误：
"error: incompatible types: int cannot be converted to String"
“错误：类型不兼容：int 无法转换为字符串”
采纳答案by SamTebbs33
Adding two chars together actually adds their unicode values and returns an integer. If you wish to instead concatenate them together as a String, then do the following:
将两个chars 加在一起实际上将它们的 unicode 值相加并返回一个整数。如果您希望将它们作为 a 连接在一起String，请执行以下操作：
return rank2 + "" + suit;
This will form a Stringand append suitto the end of rank2. So if rank2was 'J' and suitwas 'H', "JH" would be returned.
这将形成 aString并附suit加到rank2. 因此，如果rank2是 'J' 和suit'H'，则将返回“JH”。
回答by Zamrony P. Juhara
In Java, chartype is treated as int. chartype supports arithmetic operation, so they can be added. So when you use + operator, it adds their values. But because chartype and inttype are different primitive types, they are incompatible for arithmetic operation.
在 Java 中，char类型被视为int. char类型支持算术运算，因此可以将它们相加。因此，当您使用 + 运算符时，它会添加它们的值。但是因为chartype 和inttype 是不同的原始类型，所以它们对于算术运算是不兼容的。
If you need string concatenation then you need to change your code to:
如果需要字符串连接，则需要将代码更改为：
return String.valueOf(rank2) + String.valueOf(suit);
回答by Ophitect
the thing is that char can be read as an int value. When trying to perform arithmetic functions on character, it will be using the byte value of the character. Java cannot implicitly cast a int into a String since String is not regarded as a primitive data type in Java. You need to use .valueOf() method of the String class in order to explicitly tell Java that you want it as String.
问题是 char 可以被读取为 int 值。当尝试对字符执行算术函数时，它将使用字符的字节值。Java 不能将 int 隐式转换为 String，因为 String 在 Java 中不被视为原始数据类型。您需要使用 String 类的 .valueOf() 方法来明确告诉 Java 您希望它是 String。
String.valueOf(input_your_value_here);
回答by Avinash
The most efficient way for this would be following
最有效的方法是遵循
return new StringBuilder(rank2).append(suit).toString();
Here only one object is created and solves the purpose.
这里只创建了一个对象并解决了目的。
