Byte swap of a byte array into a long long
<p>I have a program where i simply copy a byte array into a long long array. There are a total of 20 bytes and so I just needed a long long of 3. The reason I copied the bytes into a long long was to make it portable on 64bit systems. </p>

<p>I just need to now byte swap before I populate that array such that the values that go into it go reversed. </p>

<p>there is a byteswap.h which has _int64 bswap_64(_int64) function that i think i can use. I was hoping for some help with the usage of that function given my long long array. would i just simply pass in the name of the long long and read it out into another long long array?
I am using c++ not .net or c#</p>

<p>update:
clearly there are issues i am still confused about. for example, workng with byte arrays that just happen to be populated with 160 bit hex string which then has to be outputed in decimal form made me think about the case where if i just do a simple assignment to a long (4 byte) array my worries would be over. Then i found out that this code would ahve to run on a 64bit sun box. Then I thought that since the sizes of data from one env to another can change just a simple assignment would not cut it. this made me think about just using a long long to just make the code sort of immune to that size issue. however, then i read about endianess and how 64bit reads MSB vs 32bit which is LSB. So, taking my data and reversing it such that it is stored in my long long as MSB was the only solution that came to mind. ofc, there is the case about the 4 extra bytes which in this case does not matter and i simply will take the decimal output and display any random six digits i choose. However programatically, i guess it would be better to just work with 4 byte longs and not deal with that whole wasted 4 byte issue. </p>
