Fast way to convert a byte[] string to its Integer value

Question

I am reading a large file from disk. That file just contains numbers, encoded as plain old ASCII. At the moment, I am reading in chunks, and then doing something like this:

byte[] token;  // bytes representing a bunch of numbers
int n = Integer.parseInt(new String(token));

In other words, I am converting to a String and then parsing the String to Integer. I would like to know if there is a way to use fast operations like shifting and binary arithmetic instead?

I suspect that this could be made faster. For example, the raw bytes for the numbers 1,2,3 are 49,50,51. Any ideas for hacks?

@Perception: no need to worry about that, I've already handled the delimination. I'm splitting out byte[] chunks as fast as possible. — Dave
– Dave, Commented Mar 6, 2013 at 9:53

irreputable · Accepted Answer · 2013-03-06 01:58:11Z

8

    int n=0;
    for(byte b : token)
        n = 10*n + (b-'0');

answered Mar 6, 2013 at 1:58

irreputable

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5 Comments

jdb Over a year ago

If you read a stream of ascii characters the order is already taken care of, right, or I am missing something?

irreputable Over a year ago

There's no endian issue here. however, will some culture write lowest digit first? Say in Arabic, we see things like عام 2013 هو عام جيد. We know Arabic is written right-to-left, so here the Arabic number is written with lowest digit first? No idea.

Dave Over a year ago

This looks quite promising! Order isn't an issue, it's left to right as expected. Also, no foreign language tricks to worry about. I'll test it in a bit and let you know :)

Dave Over a year ago

Thanks! For interest, this resulted in a roughly 28% speedup.

Achyut Over a year ago

how to convert this back to string to byte array ?

Andrew Mao · Accepted Answer · 2013-03-06 01:58:24Z

2

You can't do binary arithmetic exactly with base 10 numbers, but you can do decimal arithmetic. Assuming that higher-order digits come first:

byte[] token;
long n = 0;
long pow = 1;
for( int i = token.length - 1; i >= 0; i-- ) {
  n += (token[i]-48) * pow;
  pow *= 10;
}

answered Mar 6, 2013 at 1:58

Andrew Mao

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Comments

Evgeniy Dorofeev · Accepted Answer · 2013-03-06 05:06:44Z

0

try

    byte[] a = { 1, 2, 3 };
    for (int i = 0; i < a.length; i++) {
        a[i] += '0';
    }
    int n = Integer.parseInt(new String(a));
    System.out.println(n);

output

answered Mar 6, 2013 at 5:06

Evgeniy Dorofeev

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1 Comment

Dave Over a year ago

This is essentially the same as my original approach. I didn't want the extra overhead of creating new Strings and then parsing them to ints.

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