Can someone explain the following macro in detail?

[lilzz]

#define yahoo_put32(buf, data) ( \
(*((buf)) = (unsigned char)((data)>>24)&0xff), \
(*((buf)+1) = (unsigned char)((data)>>16)&0xff), \
(*((buf)+2) = (unsigned char)((data)>>8)&0xff), \
(*((buf)+3) = (unsigned char)(data)&0xff), \
4)


#define yahoo_get32(buf) ((((*(buf) )&0xff)<<24) + \
(((*((buf)+1))&0xff)<<16) + \
(((*((buf)+2))&0xff)<< 8) + \
(((*((buf)+3))&0xff)))

why is there a 4 at last on the yahoo_put32?

[Salem]

<< !! Posting Code? Read this First !! >>
Read it, and edit your post for readability.

[GReaper]

The first macro unpacks 4 bytes stored in 'data' to 'buf'
The second one packs those 4 bytes to make the original data

[lilzz]

The first macro unpacks 4 bytes stored in 'data' to 'buf'
The second one packs those 4 bytes to make the original data

on the first one, the unpacking, is the data's most significant 8 bits putting into the buf?

I thought the most significant 8 bits should be putting into buf+3?

and why is there's a 4 at the end?

[GReaper]

Sorry, my bad. See below.

[GReaper]

I thought the most significant 8 bits should be putting into buf+3?

Data are stored in Big Endian. It's a convertion of the web.

and why is there's a 4 at the end?

It's a comma trick. As you see, all four previous statements have a comma at the end instead of a semicolon. This essentially means the the first macro returns 4 when "executed".