vandermies - 1 year ago 110

Java Question

I need to write in a 8x8 matrix the binary values of 8 hexadecimal numbers (one for row). Those numbers will be at the most 8 bits long. I wrote the following code to convert from hexadecimal to binary:

`private String hexToBin (String hex){`

int i = Integer.parseInt(hex, 16);

String bin = Integer.toBinaryString(i);

return bin;

}

But I have the problem that values below 0x80 don't need 8 bits to be represented in binary. My question is: is there a function to convert to binary in an 8-bit format (filling the left positions with zeros)? Thanks a lot

Answer Source

Hint: use string concatenation to add the appropriate number of zeros in the appropriate place.

For example:

```
public String hexToBin(String hex) throws NumberFormatException {
String bin = Integer.toBinaryString(Integer.parseInt(hex, 16));
int len = bin.size();
return len == 8 ? bin : "00000000".substring(len - 8) + bin;
}
```

^{(Warning: untested ...)}