how to convert long to byte array in java?

Question

I have made long value from byte array using this code

byte[] by = {07, 53 -70, 74};

long value = 0;

for (int i = 0; i < by.length; i++) {
    value =  ((value << 8) + (by[i] & 0xff));
}

System.out.println(value);

out put is 520010

now I want reverse process on this and I tried it this way

long ts = 520010;

tm_stp[0] = (byte) ((byte) ts>>24);
tm_stp[1] = (byte) ((byte) ts>>16);
tm_stp[2] = (byte) ((byte) ts>>8);
tm_stp[3] = (byte) ts;
for (byte b : tm_stp) {
    System.out.println(b);
}

and output is 0 0 0 74

what is wrong in my second part of code please help me, Thanks!

A long is 64 bits in Java, so it's 8 bytes - not 4 bytes as you assume in your code. — Erwin Bolwidt, Aug 10 '16 at 08:51

Clashsoft · Accepted Answer · 2016-08-10T17:05:01.927

The problem is the fact that you cast to byte too early:

tm_stp[0] = (byte) ((byte) ts>>24);
tm_stp[1] = (byte) ((byte) ts>>16);
tm_stp[2] = (byte) ((byte) ts>>8);
                   //^^^^

This causes the ts value to be truncated, replacing the first 24 bytes with 0s. After that, shifting by anything greater than or equal to 8 will return 0. To avoid incorrect behaviour with negative values, you should also use a bitmask. The correct code should look like this:

tm_stp[0] = (byte) ((ts >> 24) & 0xFF);
tm_stp[1] = (byte) ((ts >> 16) & 0xFF);
tm_stp[2] = (byte) ((ts >>  8) & 0xFF);
tm_stp[3] = (byte) ((ts >>  0) & 0xFF); // >> 0 not actually required, only for symmetry

doing this `tm_stp[3] = (byte) ts; tm_stp[2] = (byte) (ts >> 8); tm_stp[1] = (byte) (ts >> 16); tm_stp[0] = (byte) (ts >> 24);` output is `0 7 -17 74` not actual — Jignesh Ansodariya, Aug 10 '16 at 09:19

score 1 · Answer 2 · answered Aug 10 '16 at 08:45

1

Don't convert ts to byte before shifting.

tm_stp[0] = (byte) (ts >> 24);
tm_stp[1] = (byte) (ts >> 16);
tm_stp[2] = (byte) (ts >> 8);
tm_stp[3] = (byte) ts;

answered Aug 10 '16 at 08:45

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how to convert long to byte array in java?

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