Question

该程序旨在检查任何用户字符串的第一个元素然后检查是“大写字母”还是“ 2到9之间的数字”

import java.util.*;
public class test1
{
   public static void main (String [] args)
   {
      Scanner mykey= new Scanner (System.in);
      System.out.println("Enter your first sentence");
      String firstSen= mykey.nextLine();
      String firstChar= firstSen.substring(0,1);
      if ((firstChar <='Z') && (firstChar >= 'A'))
         {System.out.println("Its a letter");}
      else if ((firstChar>='2') && (firstChar<='9'))
         {System.out.println("Its a number");}
   }
}

错误

test1.java:10: error: bad operand types for binary operator '<='
      if ((firstChar <='Z') && (firstChar >= 'A'))
                     ^
  first type:  String
  second type: char
test1.java:10: error: bad operand types for binary operator '>='
      if ((firstChar <='Z') && (firstChar >= 'A'))
                                          ^
  first type:  String
  second type: char
test1.java:12: error: bad operand types for binary operator '>='
      else if ((firstChar>='2') && (firstChar<='9'))
                         ^
  first type:  String
  second type: char
test1.java:12: error: bad operand types for binary operator '<='
      else if ((firstChar>='2') && (firstChar<='9'))

Answer 1

substring(0, 1)返回String，您的错误是使用String和char将>=与<=进行比较。

您应该使用String方法获得字符串的第一个字母，而不是char，而是charAt：

  Scanner mykey= new Scanner (System.in);
  System.out.println("Enter your first sentence");
  String firstSen= mykey.nextLine();
  char firstChar= firstSen.charAt(0); // note this line!
  if ((firstChar <='Z') && (firstChar >= 'A'))
     {System.out.println("Its a letter");}
  else if ((firstChar>='2') && (firstChar<='9'))
     {System.out.println("Its a number");}

可以将两个char与<=和>=进行比较。 String和char不能。

Answer 2

在比较两个不适合使用提供的运算符进行比较的操作数时，错误运算符的错误操作数类型出现。正如@Sweeper所指出的，您正在比较String和char。

解决问题的另一种方法是比较要比较的字符的ASCII值。这是替代解决方案。

import java.util.*;
public class Test1
{
   public static void main (String [] args)
   {
      Scanner mykey= new Scanner (System.in);
      System.out.println("Enter your first sentence");
      String firstSen= mykey.nextLine();
      // Fetch the first character and then convert to int
      // so that the comparison can be don based on ASCII values
      int firstChar = (int) firstSen.charAt(0);
      // ASCII value for A = 65, Z = 90
      if ((firstChar <= 90) && (firstChar >= 65))
         {System.out.println("Its a letter");}
      // ASCII value for 2 = 50, 9 = 57
      else if ((firstChar>= 50) && (firstChar<=57))
         {System.out.println("Its a number");}
   }
}

如何解决“错误：二进制运算符的错误操作数类型”问题？

2 个答案: