Help C++ declaration

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Help C++ declaration

 
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  #1
Mar 22nd, 2007
Hi All...
I'v a piece of code kindly help me understanding its functionality

C++ Syntax (Toggle Plain Text) 
  1.  
  2. #define FLAG_CDR 0*01 
  3. #define FLAG_MSR 0*01
  4.  
  5. unsigned Flags = AFE_REG_FLAG_CDR | AFE_REG_FLAG_MSR;
  6.  
  7.  
  8. Flags &= FLAG_MSR; //  'turn off cdr flag'
how it works...
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Re: Help C++ declaration

 
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  #2
Mar 22nd, 2007
I feel you've got something mixed up in copying these 2 lines
c Syntax (Toggle Plain Text) 
  1. #define FLAG_CDR 0*01 
  2. #define FLAG_MSR 0*01
they should probably look somethig like:
c Syntax (Toggle Plain Text) 
  1. #define FLAG_CDR 0x01 
  2. #define FLAG_MSR 0x01


This defines both flags to be same.
Anyway, just lookup the articles on bitwise operators and this should be clear. Here is another one.

This code is just trying to set some part (bits) of the value of flags variable to 0 (turning it off).
Last edited by thekashyap; Mar 22nd, 2007 at 4:04 am. Reason: Added the corrected version also.
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Re: Help C++ declaration

 
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  #3
Mar 22nd, 2007
Actually that code was like this...

C++ Syntax (Toggle Plain Text) 
  1. #define FLAG_CDR 0x01 
  2. #define FLAG_MSR 0x02
  3.  
  4. unsigned Flags = AFE_REG_FLAG_CDR | AFE_REG_FLAG_MSR;
  5.  
  6.  
  7. Flags &= FLAG_MSR; //  'turn off cdr flag'
  8. Flags &= FLAG_CDR; //  'turn off msr flag'
please tell me how the flags get off.
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Re: Help C++ declaration

 
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  #4
Mar 22nd, 2007
Did you read the articles I gave links to.. ?
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Re: Help C++ declaration

 
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  #5
Mar 22nd, 2007
yah I know the concepts of bitwise OR and AND but plz tell me that how the flags get off when we take AND of
0x01&0x02
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Re: Help C++ declaration

 
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  #6
Mar 22nd, 2007
It's hard to explain in precisely the given code. I need to know types and values of AFE_REG_FLAG_CDR and AFE_REG_FLAG_MSR. Also what're all these flags used for?

In general may be some example will help:
Say you have a class that reprents a file. An object of this would have many states. Say our object has following:
1. good - when this flag is 1 it means the object is good for usage.
2. eof - when this flag is 1 it means the file pointer is at the end of file.
3. bof - when this flag is 1 it means the file pointer is at the begenning of file.
4. error - when this flag is 1 it means that last operation performed (read/write/seek...) on file failed.
One of implementing this would be to use 4 booleans. Another is to use a single char variable (whose size say is 1 byte = 8-bits) and use 4 bits to represent these 4 flags (other 4 are ignored).
Say like this:
C++ Syntax (Toggle Plain Text) 
  1. 8    7    6    5    4    3    2    1
  2.                     ^    ^    ^    ^
  3. |<-- ignored-->|  good  err  eof  bof
Say in our class we provide 8 functions:
get_good(),set_good()
get_bad(),set_bad()
get_eof(),set_eof()
get_bof(),set_bof()
to get/set the values of these 4 flags.
Now if we had booleans: our function would look like this:
C++ Syntax (Toggle Plain Text) 
  1. bool my_class::get_good() { return m_good_bool ; }
  2. bool my_class::set_good(bool new_val) { m_good_bool = new_val;}

With bit-wise flags we do it like this:
C++ Syntax (Toggle Plain Text) 
  1. class xxx {
  2. private:
  3.     char flag ;
  4. public:
  5.     xxx() {flag = 0x00;}
  6.     bool get_good() ;
  7.     void set_good() ;
  8. };
  9.  
  10. void xxx::set_good()
  11. {
  12.     /*
  13.             0x08 = 00001000
  14.             0x00 = 00000000
  15.     0x00 | 0x08     = 00001000
  16.     and 1 | X = 1 (where X = 0 or 1)
  17.     thus irrespective of the previous value of
  18.     4th bit, it'll be set to 1.
  19.     given that 0 | X = X (where X = 0 or 1)
  20.     with this we ensure that no other flag is chagned.
  21.     */
  22.     flag |= 0x08 ;
  23. }
  24.  
  25. bool xxx::get_good()
  26. {
  27.     /*
  28.         1 & X = X (where X = 0 or 1)
  29.         0 & X = 0 (where X = 0 or 1)
  30.  
  31.             0x00 = 00000000
  32.             0x08 = 00001000
  33.     0x00 & 0x08     = 00001000
  34.  
  35.             0x08 = 00001000
  36.             0x08 = 00001000
  37.     0x00 & 0x08     = 00001000
  38.  
  39.     so if the 4th bit is set, flag & 0x08 evaluates to non-zero
  40.     if 4th bit is set, flag & 0x08 evaluates to zero.
  41.     */
  42.     return flag & 0x08 ;
  43. }
  44.  
  45.  
  46. int main()
  47. {
  48.     xxx o ;
  49.     cout << "before = " << o.get_good() << endl ;
  50.     o.set_good() ;
  51.     cout << "after = " << o.get_good() << endl ;
  52.     return 0;
  53. }
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Re: Help C++ declaration

 
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  #7
Mar 23rd, 2007
Correction for comment in get_good()

c Syntax (Toggle Plain Text) 
  1. /*
  2.     1 & X = X (where X = 0 or 1)
  3.     0 & X = 0 (where X = 0 or 1)
  4.  
  5.     0x00 = 00000000
  6.     0x08 = 00001000
  7.  0x00 & 0x08 = 00000000
  8.  
  9.     0x08 = 00001000
  10.     0x08 = 00001000
  11.  0x08 & 0x08 = 00001000
  12.  
  13.     so if the 4th bit is set, flag & 0x08 evaluates to non-zero
  14.     if 4th bit is set, flag & 0x08 evaluates to zero.
  15.     0 converted to boolean is false
  16.     non-zero converted to boolean is true
  17. */
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Re: Help C++ declaration

 
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  #8
Mar 26th, 2007
Thanks alot for the help.
I'v got the complete understanding of the topic under discussion but let me correct one thing that You quoted and that is:

C++ Syntax (Toggle Plain Text) 
  1. ...
  2. so if the 4th bit is set, flag & 0x08 evaluates to non-zero
  3. if 4th bit is set, flag & 0x08 evaluates to zero.
May I correct it as 
so if the 4th bit is set, flag & 0x08 evaluates to non-zero
if 4th bit is NOT set, flag & 0x08 evaluates to zero.
Do you agree?
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Re: Help C++ declaration

 
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  #9
Mar 26th, 2007
Yes
The 3 Laws of the Procrastination Society:
1) Never do today that which can be put off until tomorrow
2) Tomorrow never comes
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Re: Help C++ declaration

 
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  #10
Mar 27th, 2007
Yes. Copy paste err..
So now I know you've understood for sure..
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