Can someone explain the following macro in detail?

This is a discussion on Can someone explain the following macro in detail? within the C Programming forums, part of the General Programming Boards category; #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), \ ...

  1. #1
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    Can someone explain the following macro in detail?

    #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?

  2. #2
    and the hat of wrongness Salem's Avatar
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    << !! Posting Code? Read this First !! >>
    Read it, and edit your post for readability.
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    Programming Wraith GReaper's Avatar
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    The first macro unpacks 4 bytes stored in 'data' to 'buf'
    The second one packs those 4 bytes to make the original data
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    Quote Originally Posted by Sipher View Post
    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?

  5. #5
    Programming Wraith GReaper's Avatar
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    Sorry, my bad. See below.
    Last edited by GReaper; 02-23-2011 at 12:43 PM.
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  6. #6
    Programming Wraith GReaper's Avatar
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    Quote Originally Posted by lilzz View Post
    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.

    Quote Originally Posted by lilzz View Post
    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".
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