(PT) My Analysis - Question on Structures/Bit Fields

[Ravi Pandit]

Hi,    This is my analysis. The number of bytes assigned to the structure is 12 bytes and here's how.    Bits less than a byte are rounded off to the nearest Word boundary. In this case, 32 bits is the word boundary.  (sizeof(int) == 4 bytes)    int op1:5    - is assigned 1 byte  int op2:22  - is assigned 4 bytes  char d1      - is assigned 1 byte  int reg       - is assigned 4 bytes                                                       Total = 10 bytes [Data is assigned either 1 byte or in multiples of word size.] Since the architecture is 32 bits (= 4 bytes), the structure is "aligned" to a 4 byte (word) boundary.   Total = 1+4+1+4 == 10 bytes(80 bits). There will be 2 bytes[16 bits] of Padding,(again rounding off to the nearest word boundary...i.e 4*3 = 12 ..the next largest multiple of 4 that is closest to 10 is 12) .Hence the number of bytes assigned would be 12.   PS: "The word size and alignment on the machine is assumed to be 4 bytes"   Let me know any agreements/dis-agreements. Cheers -Ravi #include<stdio.h> int main() {     struct Instruct     {          int op1:5, op2:17;  // or if op2: 22 - still doesn't matter.....          char d1;          int reg;     } Instruct; printf("%d is the size of the structure",sizeof(Instruct));  return 0; }      To unsubscribe : [EMAIL PROTECTED] Yahoo! Groups Sponsor ADVERTISEMENT Yahoo! Groups Links To reply to this message, go to: http://groups.yahoo.com/group/Programmers-Town/post?act=reply&messageNum=4142 Please do not reply to this message via email. (more info)   To visit your group on the web, go to: http://groups.yahoo.com/group/Programmers-Town/   To unsubscribe from this group, send an email to: [EMAIL PROTECTED]   Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.