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First, this sounds like the problem here: How to convert a byte array to its numeric value (Java)?

But the origin of my Byte-Array is a String, something like this:

byte[] foo = new byte[8];
foo = "12345678".getBytes();

Is there a faster way (yes its really about doing this quick) than
Integer.parseInt(new String(foo))? The String contains only digits which represent a Integer.

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Karamba
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    Have you profiled your code and demonstrated that this is an actual (rather than perceived) bottleneck? – NPE Jan 12 '14 at 15:22
  • http://stackoverflow.com/questions/1030479/most-efficient-way-of-converting-string-to-integer-in-java – stinepike Jan 12 '14 at 15:28
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    Keeping the String reference, and applying Integer.parseInt to it, would save a couple of array copies. – Patricia Shanahan Jan 12 '14 at 15:29
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    It is unclear why you are using a byte array at all instead of `Integer.parseInt(originalString)`... By the way `new byte[8]` creates an array which is immediately discarded... – assylias Jan 12 '14 at 15:30
  • I am using a byte array, because the data is received via a RandomAccessFile using read(). – Karamba Jan 12 '14 at 15:36
  • I didn't profile the code, but extrapolated that it might save 2-3 seconds by multiplying the number of executions of this piece of code if the runtime could be reduced by about 50%. This is maybe a little bit too high as a expectation... – Karamba Jan 12 '14 at 15:42

2 Answers2

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try this

    int res = 0;
    for(int i = foo.length -1, m = 1; i >=0; i--, m *= 10) {
        res += (foo[i] & 0xF) * m; 
    }
Evgeniy Dorofeev
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1

You could try something like this:

byte foo[] = "12345678".getBytes();
//Since it is an 'integer' essentially, it will contain ASCII values of decimal digits.
long num = 0;  //Store number here.
for(int i = foo.length - 1; i >= 0; i--)
{
    num = num * 10 + (foo[i] - '0'); // or (foo[i] - 48) or (foo[i] & 0xf)
}

num stores the required number.

Precaution: Make sure you use decimal number only.

EDIT:

The Mechanism

On calling getBytes() of the String "12345678", the byte[] returned is as follows:

enter image description here

The values we see are the ASCII or Unicode values for the eqivalent characters. There are several ways to extract their equivalent character as ints:

  1. Since the arrangement of the digit chars, i.e. '0', '1', '2', etc. are done in the desired order - ascending and sequentially, we can extract the characters by subtrcting the ASCII value of '0' i.e. 48.
  2. @Evgeniy Dorofeev correctly pointed out the method of masking:

'0' => 48 => 11 0000

We notice that if we extract the last 4 bits, we get the required int. To do this, we need to extract them in the following way. Let us take foo[1], i.e. 50

  50      & 0xf  (original)
= 50      & 15   (in Decimal)
= 11 0010 & 1111 (in Binary)
= 0010           (result)
= 2              (Decimal)

Hence, the required digit is obtained. It in necessary to add it to num int the correct way (which I expect of every programmer to have some knowledge about).

Hungry Blue Dev
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