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bit structure for two dimmensional arrays

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mformazhar1980

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bit structure for arrays

hi

i am intending to make a bit structure that i am going to use for storing boolean data into a two dimensional array.
the problem is i dont know exactly which data type is suitable so that i could save memory. i have used uint8_t as a data type. all i need to check whether i am on right track or not.

i want to store bits i.e 1's and 0's in a two dimensional array.
is that really save only the bits in said array or is it saving the whole bit in the index?
i.e suppose Value[0][j] stores {1,1,1,1,1}or
Value[1][j] stores {0,0,0,0,0}

the code snippet is as follows
Code:
unit8_t Value[4][10];

typedef struct Save_Bits
{
uint8_t  i  :1;
uint8_t  j  :1;

} Save_Bits;

struct Save_Bits  Value[i][j];



// then in the while loop get values

now the point is that how i can fetch all the data so that it can represent like

Value[0][j] stores all the 0's untill there is 1's appear and then it should immediately switches to Value[1][j] and store all the 1's there until the next change i.e occurrence of 0's again. and switches to Value[2][j] and store all the 0,s there and so on.
can any body help me to sort out the logic.
thanks


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btbass

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Re: bit structure for arrays

One possible way to do it would be in a single array of bit counts.

psuedo code:

Code:
uint8_t data[ARRAY_SIZE];
uint8_t index = 0;
uint8_t count = 0;

while(index < ARRAY_SIZE)
  {
  while(READ_DATA() == 1)  /* Function READ_DATA() returns boolean */  
    {
    data[index] = ++count;
    }

  index++;
  data[index] = count = 1;

  while(READ_DATA() == 0)
    {
    data[index] = ++count;
    }

  index++;
  data[index] = count = 1;
  }

You could reverse the logic to output the results.
 

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