我如何从ASCII转换为十进制

Question

I have this code:

final byte[] bslval = data1.getByteArray(HRPService.BSL_VALUE);

runOnUiThread(new Runnable() 
{
    public void run() 
    {
        if (bslval != null) 
        {
            try 
            {
                Log.i(TAG, "BYTE BSL VAL =" + bslval[0]);
                //TextView bsltv = (TextView) findViewById(R.id.BodySensorLocation);
                EditText bsltv = (EditText) findViewById(R.id.EditSensorLocation);
                String bslstr = new String(bslval, "UTF-8");
                bsltv.setText(bslstr);                                
            } 
            catch (Exception e) 
            {
                Log.e(TAG, e.toString());
            }

The EditText does show the value as a "d" but not as "64" (which I want). How do I convert this to a decimal instead?

Answer 1

Something like this:

String bslstr = String.valueOf(bslval[0] & 0xff);

This assumes, the value is actually in bslval[0] . The & 0xff is there to get positive results for byte values between (byte)0x80 and (byte)0xff .

Answer 2

如果您只想将第一个字节值bslval[0]为文本，请使用更简单的字符串连接。

bsltv.setText(bslval[0] + "");