If you absolutely need to get at the 32-bit integer value, you can use

union mytype {
{ struct bits {
    unsigned int a:6;
    unsigned int b:4;
    unsigned int c:8;
    unsigned int d:5;
    unsigned int e:9;
  } b;
  unsigned int v;
} variable;

then access the bit-fields like

  unsigned int x = variable.b.e;
  variable.b.a = 3;
  x = variable.v;


  


On 08/28/2009 12:58 -0700, Ben Rosenberg wrote:
>>	On Fri, Aug 28, 2009 at 11:40 AM, Randi Botse<nightdecoder@gmail.com> wrote:
>>	> Ben, after assign all bitfield struct member, how to pack them to be a
>>	> 32bit integer value?
>>	>
>>	> DEC      BIN
>>	> 43        101011        -> foo.a (6 bit)
>>	> 11        1011            -> foo.b (4 bit)
>>	> 120      01111000    -> foo.c (8 bit)
>>	> 30        11110          -> foo.d (5 bit)
>>	> 418      110100010  -> foo.e (9 bit)
>>	>
>>	> the bit pattern is : 10101110110111100011110110100010 or 2933800354 in
>>	> decimal, is this possible?
>>	>
>>	> while with masking and shift i can pack them with:
>>	> ((foo.a & 0x3f) << 26) | ((foo.b & 0xf) << 22) | ((foo,c & 0xff) <<
>>	> 14) | ((foo.d & 0x1f) << 9) | (foo.e & 0x1ff)
>>	>
>>	>
>>	
>>	The bitfield struct is already packed, you don't need to do anything
>>	else. Just assign values to the different parts.
>>	
>>	Ben
>>	--
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End of included message



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