我正在编写一个在十进制、二进制和十六进制之间转换的程序。程序会编译,但是当我输入二进制到十六进制时,我会得到一个异常"java.lang.StringIndexOutOfBoundsException: String超出范围: 34“。这是第一个问题。第二个问题是十六进制到二进制转换给了我一些非常长(而且不正确)的返回。我已经把两者的密码都包括进去了。向正确的方向前进是非常值得赞赏的。
二进制到十六进制:
/**
* Method that converts a binary number to its hexadecimal equivalent.
* @param no parameters
* @return returns void
*/
public void binToHex()
{
System.out.println("The binary number you enter will be converted to its hexidecimal equivalent.");
System.out.println("Please enter a binary number: ");
Scanner keyboard = new Scanner(System.in);
String bin = keyboard.nextLine();
String oldbin = bin;
bin = bin.replace(" ", "").trim();
StringBuffer hex = new StringBuffer("00000000000000000000000000000000");
//String hex1 = "";
int j = 0;
for (int i = 0; i < bin.length(); i++)
{
if (bin.substring(i, i+4).equals("0000"))
{
hex.setCharAt(j, '0');
}
else if (bin.substring(i, i+4).equals("0001"))
{
hex.setCharAt(j, '1');
}
else if (bin.substring(i, i+4).equals("0010"))
{
hex.setCharAt(j, '2');
}
else if (bin.substring(i, i+4).equals("0011"))
{
hex.setCharAt(j, '3');
}
else if (bin.substring(i, i+4).equals("0100"))
{
hex.setCharAt(j, '4');
}
else if (bin.substring(i, i+4).equals("0101"))
{
hex.setCharAt(j, '5');
}
else if (bin.substring(i, i+4).equals("0110"))
{
hex.setCharAt(j, '6');
}
else if (bin.substring(i, i+4).equals("0111"))
{
hex.setCharAt(j, '7');
}
else if (bin.substring(i, i+4).equals("1000"))
{
hex.setCharAt(j, '8');
}
else if (bin.substring(i, i+4).equals("1001"))
{
hex.setCharAt(j, '9');
}
else if (bin.substring(i, i+4).equals("1010"))
{
hex.setCharAt(j, 'A');
}
else if (bin.substring(i, i+4).equals("1011"))
{
hex.setCharAt(j, 'B');
}
else if (bin.substring(i, i+4).equals("1100"))
{
hex.setCharAt(j, 'C');
}
else if (bin.substring(i, i+4).equals("1101"))
{
hex.setCharAt(j, 'D');
}
else if (bin.substring(i, i+4).equals("1110"))
{
hex.setCharAt(j, 'E');
}
else if(bin.substring(i, i+4).equals("1111"))
{
hex.setCharAt(j, 'F');
}
i = i + 4;
j = j + 1;
}
System.out.println("The binary number you entered, " + oldbin + " is " + hex + " in hexadecimal.\n");
pw.print("The binary number you entered, " + oldbin + " is " + hex + " in hexadecimal.\n");
}}
十六进制到二进制:
/**
* Method that converts a hexadecimal number to its binary equivalent.
* @param no parameters
* @return returns void
*/
public void hexToBin()
{
System.out.println("The hexadecimal number you enter will be convered to its binary equivalent.");
System.out.println("Please enter a hexadecimal number: ");
Scanner keyboard = new Scanner(System.in);
String bin = keyboard.nextLine();
bin = bin.trim();
String binary = "";
for (int i = 0; i < bin.length(); i++)
{
if(bin.charAt(i) == '0')
{
binary = binary.concat("0000");
}
else if(bin.charAt(i) == '1')
{
binary = binary.concat("0001");
}
else if(bin.charAt(i) == '2')
{
binary = binary.concat("0010");
}
else if(bin.charAt(i) == '3')
{
binary = binary.concat("0011");
}
else if(bin.charAt(i) == '4')
{
binary = binary.concat("0100");
}
else if(bin.charAt(i) == '5')
{
binary = binary.concat("0101");
}
else if(bin.charAt(i) == '6')
{
binary = binary.concat("0110");
}
else if(bin.charAt(i) == '7')
{
binary = binary.concat("0111");
}
else if(bin.charAt(i) == '8')
{
binary = binary.concat("1000");
}
else if(bin.charAt(i) == '9')
{
binary = binary.concat("1001");
}
else if(bin.charAt(i) == 'A');
{
binary = binary.concat("1010");
}
if(bin.charAt(i) == 'B');
{
binary = binary.concat("1011");
}
if(bin.charAt(i) == 'C');
{
binary = binary.concat("1100");
}
if(bin.charAt(i) == 'D');
{
binary = binary.concat("1101");
}
if(bin.charAt(i) == 'E');
{
binary = binary.concat("1110");
}
if(bin.charAt(i) == 'F');
{
binary = binary.concat("1111");
}
}
System.out.println("The hexadecimal you entered, " + bin + " is " + binary + " in binary.\n");
pw.print("The hexadecimal you entered, " + bin + " is " + binary + " in binary.\n");
}}
发布于 2014-09-05 19:29:19
在binToHex中:
第一个问题是,在bin.substring(i, i+4)中,i+4可能超出了界限,因为i和bin.length()-1一样高。
THe第二个问题是,您不知道二进制字符串是可以除数4的。您应该将它与零放在一起。
完成左填充后,可以将循环更改为:
for (int i = 0; i < bin.length(); i+=4)
那么bin.substring(i, i+4)就永远不会超出界限。
编辑:我刚刚注意到,每次迭代结束时,我都会增加4。但是,您还在for循环中增加了1,因此在每次迭代中总共增加了5。
我还建议您使用StringBuilder而不是StringBuffer (您不需要线程安全)。并使用append方法向其添加字符。不要在第二种方法中使用String.concat。使用StringBuilder。
在hexToBin中:
在某些条件下,你把else忘在ifs之前,你已经";“了。
例如:if(bin.charAt(i) == 'B');
这意味着遵循该条件的代码将始终被执行。
正确的实现应该是(尽管我仍然建议使用StringBuilder ):
if(bin.charAt(i) == '0')
{
binary = binary.concat("0000");
}
else if(bin.charAt(i) == '1')
{
binary = binary.concat("0001");
}
else if(bin.charAt(i) == '2')
{
binary = binary.concat("0010");
}
else if(bin.charAt(i) == '3')
{
binary = binary.concat("0011");
}
else if(bin.charAt(i) == '4')
{
binary = binary.concat("0100");
}
else if(bin.charAt(i) == '5')
{
binary = binary.concat("0101");
}
else if(bin.charAt(i) == '6')
{
binary = binary.concat("0110");
}
else if(bin.charAt(i) == '7')
{
binary = binary.concat("0111");
}
else if(bin.charAt(i) == '8')
{
binary = binary.concat("1000");
}
else if(bin.charAt(i) == '9')
{
binary = binary.concat("1001");
}
else if(bin.charAt(i) == 'A')
{
binary = binary.concat("1010");
}
else if(bin.charAt(i) == 'B')
{
binary = binary.concat("1011");
}
else if(bin.charAt(i) == 'C')
{
binary = binary.concat("1100");
}
else if(bin.charAt(i) == 'D')
{
binary = binary.concat("1101");
}
else if(bin.charAt(i) == 'E')
{
binary = binary.concat("1110");
}
else if(bin.charAt(i) == 'F')
{
binary = binary.concat("1111");
}https://stackoverflow.com/questions/25692730
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