Does memcpy cause endian flip?

[Subsonics]

Am I doing something wrong here or what is it that cause the number to come out backwards?

	unsigned char bytes[3];	
	unsigned short large = 0x87E5; 

	printf("%x\n", large);
	
	memcpy(&bytes, &large, 2);
	printf("%x%x\n", bytes[0], bytes[1]);

This is the result:

87e5
e587

[Dino]

Your PC is a little endian machine. That's simply the way it is stored. In other words, everything is working as designed.

[Subsonics]

Hmm, ok thats backwards as far as I as a human see thigs but ok. Suppose that I need to do bit manipulations on the first value would I need to use the htons function on it first for it to come out right?

[Dino]

If you do bit operations on a short, as a short, the system will work as you expect.
If you do bit operations on a char, as a char, the system will work as you expect.
... etc for all datatypes.

If you do bit operations on a short as a char, or vice versa, you will have to deal with the hardware implementation (big or little endianness) of the platform you are programming on.

[Subsonics]

If you do bit operations on a short as a char..

Do you mean if I break up a short into two chars as above?

[Dino]

Yes.

[Subsonics]

Thanks,

Is there a way to print a number as binary with printf BTW?

[Dino]

Not with a built-in formatter. I bet you're smart enough to be able to write a loop to do it though.