Question

I need to alloc an array of uint64_t[1e9] to count something, and I know the items are between (0,2^39).
So I want to calloc 5*1e9 bytes for the array.

Then I found that, if I want to make the uint64_t meanful, it is difficult to by pass the byte order.

There should be 2 ways.

One is to check the endianness first, so that we can memcpy the 5 bytes to either first or last of the whole 8 bytes.

The other is to use 5 bitshift and then bit-or them together.

I think the former should be faster.

So, under GCC or libc or GNU system, is there any header file to indicate whether the current system is Little Endian or Big Endian ? I know x86_64 is Little Endian, but I don’t like to write a unportable code.

Of course any other idears are welcomed.

Add:

I need use the array to count many strings use D-left hashing. I plan to use 21bit for key and 18bit for counting.

Answer 1

Well, I just find the kernel headers come with <asm/byteorder.h>.

inline memcpy to a while(i<x+3){++*i=++*j} may still slower since cache operation is slower than registers.

another way for memcpy is:

union dat {
 uint64_t a;
 char b[8];
} d;