views:

159

answers:

4

I'm looking through some code for learning purposes. I'm working through this portion of code.

// e.g. const unsigned char data={0x1,0x7C ... }
unsigned char buf[40];
memset(buf,0,40);
buf[0] = 0x52;
memcpy(buf+1, data, length); // What does buf+1 do in this situation?

On the last line where memcpy is called what does buf+1 do? buf is a character array, so what does +1 do to it?

+3  A: 

buf+1 is the same as &(buf[1]). In other words, it returns a pointer to the 2nd (index 1) character of buf.

derobert
Only 29 seconds too slow :P
rpetrich
+5  A: 

buf+1 is equivalent to &(buf[1])

rpetrich
+4  A: 

In C, every array name is a pointer, so buf here also means the pointer which point to buf[0].Then "buf+1" means "buf[1]"'s address.

zxeoc
Ah cool. I had a hunch that was what it was doing, but the syntax did make sense to me. Thanks for clearing that up for me!
macinjosh
caf