问题描述
主要方法:
public static void main(String args[])
{
String inputByUserString=""; //number input by the player
int inputByUserInteger;
Scanner sc=new Scanner(system.in);
System.out.println("Enter a number"); //getting the input
inputByUserString=sc.nextLine();
inputByUserInteger=Integer.parseInt(inputByUserString);
然后创建一个切换用例,其中有更多的转换选项,例如十进制数到二进制数等...
int binaryToDecimalNumberVariable=obj1.binaryToDecimalConversion(inputByUserInteger);
System.out.println("Binary Number:"+inputByUserInteger);
System.out.println("Decimal Number:"+binaryToDecimalNumberVariable);
从二进制到十进制的转换方法:
public int binaryToDecimalConversion(int inp) //to convert binary number into decimal number
{
int a1;int a2=0;int a3 = 0;
while(inp>0)
{
a1=inp%10;
a1=inp/10;
a3=(int)(a3+(a1*(Math.pow(2,a2)))); //result
a2++;
}
return a3;
}
整个开关盒:
System.out.println("What type of conversion do you want?");
System.out.println("1)Decimal number to Binary number");
System.out.println("2)Binary number to Decimal number");
System.out.println("3)Decimal number to Octal number");
System.out.println("4)Octal number to Decimal number");
System.out.println("5)Decimal number to Hexadecimal number");
System.out.println("6)Hexadecimal number to Decimal number");
System.out.println("7)Octal number to Hexadecimal number");
System.out.println("8)Hexadecimal number to Octal number");
int choice=sc.nextInt();
switch(choice)
{
case 1: //Decimal number to Binary number
break;
case 2: //Binary number to Decimal number
int binaryToDeci=obj1.binaryToDecimalConversion(inputByUserInteger);
System.out.println("Binary Number:"+inputByUserInteger);
System.out.println("Decimal Number:"+binaryToDeci);
break;
case 3: //Decimal number to Octal number
break;
case 4: //Octal number to Decimal number
break;
case 5: //Decimal number to Hexadecimal number
break;
case 6: //Hexadecimal number to Decimal number
break;
case 7: //Octal number to Hexadecimal number
break;
case 8: //Hexadecimal number to Octal number
break;
default:
System.out.println("Invalid Input");
} //switch close
所以,请帮帮我。
解决方法
要进一步说明我的观点,请参阅处理您的要求的正确方法:
String myBinaryNr = sc.next();
int myNr = Integer.parseInt(myBinaryNr,2);
String myDecimalNr = Integer.toString(myNr,10);
后10个是可选的,因为它是默认设置。
,如果卡住,则可能尚未检测到输入,正在等待输入。为了确认这一点,我将添加一个
System.out.println("Test" + choice);
就在选择输入下,所以会像这样:
int choice = sc.nextInt();
System.out.println("Test" + choice);
switch(choice){...}
如果无法正确打印,则表明您已找到问题。它可能与System.in
没关系,我只需要这样做就可以解决问题。
public int binaryToDecimalConversion(int inp) //to convert binary number into decimal
{
int a1;int a2=0;int a3 = 0;
while(inp>0)
{
a1=inp%10;
a3=(int)(a3+(a1*(Math.pow(2,a2)))); //result
a2++;
inp=inp/10;
}
return a3;
}