在我正在开发的应用程序中,首先,我将十进制值转换为十六进制字符串。
例如,我将 int 值 -100000 转换为十六进制值:
hex_value = Integer.toHexString (-100000)
我明白了:
hex_value = FFFE7960
现在,我需要进行逆转换。我得到 FFFE7960 十六进制值,我需要再次将其转换为 -100000,为此我使用:
int_value = Integer.parseInt(hex_value,16)
但是我得到的不是-100000,而是4294867296。
因此,我得到的不是有符号的 int 值,而是无符号的值,这会导致我的应用程序出现错误。
如何才能获得所需的值?
更新 - 完整代码
这是我通过蓝牙收到的字符串:
s = "2b 00 ff fe 79 60"
使用 StringTokenizer 我将其拆分:
StringTokenizer tokens = new StringTokenizer(s," ");
String one = tokens.nextToken();
String two = tokens.nextToken();
String three = tokens.nextToken();
String four = tokens.nextToken();
String five = tokens.nextToken();
String six = tokens.nextToken();
received_hexValue = three + four + five + six;
所以,received_hexValue = "fffe7960"
现在,我进行所需的转换:
int_value_receive = (int)Long.parseLong(received_hexValue, 16);
int_value_receive = -200000 - int_value_receive;
newIntValue = (int_value_receive * 100) / (200000 * (-1));
在第一行中,当从十六进制转换为整数时,调试器抛出一个 Long.invalidLong(String) 错误,并在 logcat 中显示注释的错误:java.lang.NullPointerException
最佳答案
尝试使用以下代码。
String hex_value = Integer.toHexString(-100000);
Log.d("Home", "Hex : " + hex_value);
int int_value = (int) Long.parseLong(hex_value, 16);
Log.d("Home", "Int : " + int_value);
此代码将首先创建长值 4294867296
,然后返回 -100000
作为输出。
编辑
你的代码就像
String s = "2b 00 ff fe 79 60";
StringTokenizer tokens = new StringTokenizer(s, " ");
String one = tokens.nextToken();
String two = tokens.nextToken();
String three = tokens.nextToken();
String four = tokens.nextToken();
String five = tokens.nextToken();
String six = tokens.nextToken();
String received_hexValue = three + four + five + six;
Log.d("Home", "Hex : " + received_hexValue);
//to get sub string of your length, pass start and end offset
Log.d("Home", "SubString Hex : " +received_hexValue.substring(0, 8));
int int_value_receive = (int) Long.parseLong(received_hexValue, 16);
Log.d("Home", "Old Int : " + int_value_receive);
int_value_receive = -200000 - int_value_receive;
Log.d("Home", "Int : " + int_value_receive);
int newIntValue = (int_value_receive * 100) / (200000 * (-1));
Log.d("Home", "New Int : " + newIntValue);
输出
09-03 16:22:37.421: 调试/Home(28973): 十六进制: fffe7960
09-03 16:22:37.421:调试/主页(28973):旧整数:-100000
09-03 16:22:37.421: 调试/主页(28973): Int : -100000
09-03 16:22:37.421:调试/主页(28973):新整数:50
关于android - 从十六进制字符串到有符号整数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18589357/