Java 错误:类中的构造函数无法应用于给定类型
编写一个代码,将字符串作为电话号码,并用字母代替数字,然后将该字符串解码为数字。当我编写tester类时,它告诉我来自另一个类的构造函数不能应用于测试用例中的给定类型 这是代码的主要部分Java 错误:类中的构造函数无法应用于给定类型,java,string,Java,String,编写一个代码,将字符串作为电话号码,并用字母代替数字,然后将该字符串解码为数字。当我编写tester类时,它告诉我来自另一个类的构造函数不能应用于测试用例中的给定类型 这是代码的主要部分 public class PhoneNumber { String phoneNumber; String telNumber; public PhoneNumber(String num, String telNum) { phoneNumber = num;
public class PhoneNumber
{
String phoneNumber;
String telNumber;
public PhoneNumber(String num, String telNum)
{
phoneNumber = num;
telNumber = telNum;
}
public String decodePhoneNumber()
{
//Takes string form phone number and decodes based on number pad
//Find code that makes if statement not care about caps
//so if a || b || c number[cnt] = 1 etc..
for (int cnt = 0; cnt < phoneNumber.length(); cnt++)
{
char num = phoneNumber.toLowerCase().charAt(cnt);
if ((num == 'a') || (num == 'b') || (num == 'c'))
{
//number is 2
telNumber = telNumber + "2";
}
else if ((num == 'd') || (num == 'e') || (num == 'f'))
{
//number is 3
telNumber = telNumber +"3";
}
else if ((num == 'g') || (num == 'h') || (num == 'i'))
{
//number is 4
telNumber = telNumber +"4";
}
else if ((num == 'j') || (num == 'k') || (num == 'l'))
{
//number is 5
telNumber = telNumber +"5";
}
else if ((num == 'm') || (num == 'n') || (num == 'o'))
{
//number is 6
telNumber = telNumber +"6";
}
else if ((num == 'p') || (num == 'q') || (num == 'r') || (num == 's'))
{
//number is 7
telNumber = telNumber +"7";
}
else if ((num == 't') || (num == 'u') || (num == 'v'))
{
//number is 8
telNumber = telNumber +"8";
}
else
{
//number is 9
telNumber = telNumber +"9";
}
}
return telNumber;
}
}
错误来自测试类的第5行。
PhoneNumber
类构造函数有两个字符串参数public PhoneNumber(string num,string telNum)
传递另一个字符串参数或使用一个字符串参数创建另一个构造函数
像这样初始化对象
PhoneNumber ph1 = new PhoneNumber("XXXXX","1-800-ILOVENY");
或者在PhoneNumber
类中创建一个字符串参数的另一个构造函数
public PhoneNumber( String telNum)
{
telNumber = telNum;
}
你的
PhoneNumber
唯一的构造函数接受两个字符串参数。哇,我的疏忽很简单。今天我在看几个不同的项目,所以这是漫长的一天。真不敢相信我错过了。
public PhoneNumber( String telNum)
{
telNumber = telNum;
}