我遇到了砖墙,我不知道如何继续。
快速总结:我的代码假设将 4 个十六进制字符串(ramError1-4)手动转换为二进制。我希望能够循环 4 次,并且每次创建一个唯一的 StringBuilder 以继续下一步。
在过去的两天里,我一遍又一遍地尝试让它工作但没有成功。任何意见,将不胜感激。这是我没有失败循环的代码。Hex2 作为测试工作,但我希望能够循环通过 HexCodes 1-4(ramError1-4)
那么如何在创建 StringBuilder 时循环 Hex1-4 呢?
注意:ramErrors1-4 中确实包含我之前通过读取文件在代码中添加的字符串。
String Hex1=ramError1;
String Hex2=ramError2;
String Hex3=ramError3;
String Hex4=ramError4;
StringBuilder hexstring1 = new StringBuilder();
StringBuilder hexstring2 = new StringBuilder();
StringBuilder hexstring3 = new StringBuilder();
StringBuilder hexstring4 = new StringBuilder();
for (int x = 0; x <= 8; x++)
{
if (Hex2.charAt(x) == 'A')
{
hexstring1.append("1010");
}
else if (Hex2.charAt(x) == 'B')
{
hexstring1.append("1011");
}
else if (Hex2.charAt(x) == 'C')
{
hexstring1.append("1100");
}
else if (Hex2.charAt(x) == 'D')
{
hexstring1.append("1101");
}
else if (Hex2.charAt(x) == 'E')
{
hexstring1.append("1110");
}
else if (Hex2.charAt(x) == 'F')
{
hexstring1.append("1111");
}
else if (Hex2.charAt(x) == '0')
{
hexstring1.append("0000");
}
else if (Hex2.charAt(x) == '1')
{
hexstring1.append("0001");
}
else if (Hex2.charAt(x) == '2')
{
hexstring1.append("0010");
}
else if (Hex2.charAt(x) == '3')
{
hexstring1.append("0011");
}
else if (Hex2.charAt(x) == '4')
{
hexstring1.append("0100");
}
else if (Hex2.charAt(x) == '5')
{
hexstring1.append("0101");
}
else if (Hex2.charAt(x) == '6')
{
hexstring1.append("0110");
}
else if (Hex2.charAt(x) == '7')
{
hexstring1.append("0111");
}
else if (Hex2.charAt(x) == '8')
{
hexstring1.append("1000");
}
else if (Hex2.charAt(x) == '9')
{
hexstring1.append("1001");
}
else
{
System.out.println("error at char" + x );
}
}
System.out.println("Hex To Binary is " + hexstring1.toString());
System.out.println("Hex To Binary is " + hexstring2.toString());
System.out.println("Hex To Binary is " + hexstring3.toString());
System.out.println("Hex To Binary is " + hexstring4.toString());
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