Thread: erase

BigInteger PowerOfTwo = BigInteger.valueOf(2);
          DecimalToBinary = new BigInteger(LargeBase.getText());
          String binaryString = DecimalToBinary.toString(2);
          int binaryNo = Integer.parseInt(binaryString, 2); 
	  String backTOString = Integer.toBinaryString(binaryNo);
          char[] binaryArray = backTOString.toCharArray();
          int FristCount=0;int SecCount=0;
          for (char c : binaryArray){
              if(c == 49){ // int i=c; will convter to ascii, so 1=49, 0=48.
 
                  display.setText(display.getText()+" "+PowerOfTwo+"^"+FristCount+" + ");}
                  FristCount++;
      }

this code is working, however the print out want to be 2+2.
However the print out come as 2+2+.
How to earse the last part of +
so come it will cone out like 2+2.

It might help if you explained what the code is supposed to be doing...

display.setText(display.getText()+" "+PowerOfTwo+"^"+FristCount+" + ");}

Is this the code that adds the ending +?
You need change the logic to put the + at the beginning of the concatenation vs at the end as you have here.

Your code will always print an extra " + ", you can't just erase it.
You can fix this by putting in an if() statement to check if there is more numbers to output.

if(c == 49){ // int i=c; will convter to ascii, so 1=49, 0=48.

display.setText(display.getText()+" "+PowerOfTwo+"^"+FristCount);
if( //more to print)
display(" + ");
}

FristCount++;

Why not use:
if(c == '1')

Originally Posted by Norm

Why not use:
if(c == '1')

c is char, when char convert to int. it will change to 49, when char is 1.

Originally Posted by firebluetom

Your code will always print an extra " + ", you can't just erase it.
You can fix this by putting in an if() statement to check if there is more numbers to output.

if(c == 49){ // int i=c; will convter to ascii, so 1=49, 0=48.

display.setText(display.getText()+" "+PowerOfTwo+"^"+FristCount);
if( //more to print)
display(" + ");
}

FristCount++;

how to use if statement to check?