我是一个自我学习者。我想将角色转换为相应的二进制文件。例如,我给出了角色' A' (它的ASCII为65,二进制为65为1000001)我想得到答案为1000001。 提前谢谢..
这是我的代码
import android.app.AlertDialog;
import android.os.Bundle;
import android.support.design.widget.FloatingActionButton;
import android.support.design.widget.Snackbar;
import android.support.v7.app.AppCompatActivity;
import android.support.v7.widget.Toolbar;
import android.text.Editable;
import android.text.TextWatcher;
import android.view.View;
import android.view.Menu;
import android.view.MenuItem;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import java.util.Stack;
public class MainActivity extends AppCompatActivity implements TextWatcher,View.OnClickListener
{
EditText txtDecimal;
TextView txtBinary;
Button btnAbout;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
txtDecimal=(EditText)findViewById(R.id.txtDecimal);
txtBinary=(TextView)findViewById(R.id.txtval2);
//txtdec=(TextView)findViewById(R.id.txtDecimal);
txtDecimal.addTextChangedListener(this);
btnAbout=(Button)findViewById(R.id.button1);
btnAbout.setOnClickListener(this);
}
public void beforeTextChanged(CharSequence sequence,int start,int count,int after)
{
}
public void afterTextChanged(Editable editable)
{
}
public void onTextChanged(CharSequence sequence,int start,int before,int count) {
calculate(2, txtBinary); // for base 2 (binary)
}
public void calculate(int base,TextView txtView)
{
if(txtDecimal.getText().toString().trim().length()==0)
{
txtView.setText("");
return;
}
try
{
Stack<Object> stack=new Stack<Object>();
int number=Integer.parseInt(txtDecimal.getText().toString());
while (number>0)
{
int remainder=number%base; // find remainder
if(remainder<10)
// for remainder smaller than 10
{
stack.push(remainder);
// push remainder in stack
}
number/=base;
}
StringBuffer buffer=new StringBuffer();
while (!stack.isEmpty())
{
buffer.append(stack.pop().toString());
}
txtView.setText(buffer.toString());
}
catch (Exception e)
{
txtView.setText(e.getMessage());
}
}
public void onClick(View view)
// to display Information in a dialog box
{
// create a dialog box
AlertDialog.Builder builder=new AlertDialog.Builder(this);
// to allow cancelling the dialog box
builder.setCancelable(true);
// set title
builder.setTitle("About NumberSystemConverter");
// set message
builder.setMessage("Made by HARI");
// display dialog box
builder.show();
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.menu_main, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
// Handle action bar item clicks here. The action bar will
// automatically handle clicks on the Home/Up button, so long
// as you specify a parent activity in AndroidManifest.xml.
int id = item.getItemId();
//noinspection SimplifiableIfStatement
if (id == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
}
通过这段代码,我只能转换数字。我也想转换字符。通过这段代码,如果我,我给它,它告诉我&#34;无效的一个&#34;。我想同时处理两个字符和numbers.plz帮助我...
答案 0 :(得分:2)
尝试以下
String s = txtDecimal.getText().toString();
String binary_num ="";
StringBuilder sb = new StringBuilder();
String ascString = null;
long asciiInt;
for (int i = 0; i < s.length(); i++){
sb.append((int)s.charAt(i));
char c = s.charAt(i);
}
ascString = sb.toString();
asciiInt = Long.parseLong(ascString);
ascii_number =(int) intasciiInt;
int binary[] = new int[ascii_number];
int index = 0;
while(ascii_number > 0){
binary[index++] = ascii_number%2;
ascii_number = ascii_number/2;
}
for(int i = index-1;i >= 0;i--){
binary_num = binary_num + binary[i];
}
txtBinary.setText(binary_num);
答案 1 :(得分:0)
通过此代码,我只能转换数字。
这是不正确的。该代码还可以将字符转换为“二进制”。只需更换:
int number=Integer.parseInt(txtDecimal.getText().toString());
带
int number = (int) someCharacter;
它将转换它。
话虽如此,您的代码根本没有转换为二进制文件。它实际上是转换为表示为字符串的base-2数字序列。如果你使用大于10的calculate) with
base`,那么该字符串将不能作为base-N数字解码。