How do I get bit-by-bit data from an integer value in C

Question

I want to extract bits of a decimal number   For example  7 is binary 0111  and I want to get 0 1 1 1  all bits stored in bool  How can I do so   OK  a loop is not a good option  can I do something else for this

User · Accepted Answer

If you want the k-th bit of n  then do    n  amp    1  lt  lt  k     gt  gt  k   Here we create a mask  apply the mask to n  and then right shift the masked value to get just the bit we want  We could write it out more fully as       int mask    1  lt  lt  k      int masked n   n  amp  mask      int thebit   masked n  gt  gt  k    You can read more about bit-masking here    Here is a program    include  lt stdio h gt   include  lt stdlib h gt   int  get bits int n  int bitswanted     int  bits   malloc sizeof int    bitswanted      int k    for k 0  k lt bitswanted  k         int mask    1  lt  lt  k      int masked n   n  amp  mask      int thebit   masked n  gt  gt  k      bits k    thebit         return bits     int main      int n 7     int  bitswanted   5     int  bits   get bits n  bitswanted      printf   d      n      int i    for i bitswanted-1  i gt  0 i--       printf   d    bits i           printf   n

User · Answer

include  lt stdio h gt   int main void        int number   7     signed        int vbool 8   sizeof int        int i          for  i   0  i  lt  8   sizeof int   i                          vbool i    number lt  lt i  lt  0                 printf   d   vbool i                  return 0

User · Answer

Using std  bitset  int value   123  std  bitset lt sizeof int  gt  bits value   std  cout  lt  lt bits to string

User · Answer

Here s a very simple way to do it   int main         int s 7 l 1      vector  lt bool gt  v      v clear        while  l  lt   4                v push back s 2           s    2          l              for  l  v size  -1   l  gt   0  l--                cout lt  lt v l  lt  lt                return 0

User · Answer

As requested  I decided to extend my comment on forefinger s answer to a full-fledged answer  Although his answer is correct  it is needlessly complex  Furthermore all current answers use signed ints to represent the values  This is dangerous  as right-shifting of negative values is implementation-defined  i e  not portable  and left-shifting can lead to undefined behavior  see this question    By right-shifting the desired bit into the least significant bit position  masking can be done with 1  No need to compute a new mask value for each bit    n  gt  gt  k   amp  1   As a complete program  computing  and subsequently printing  an array of single bit values    include  lt stdio h gt   include  lt stdlib h gt   int main int argc  char   argv        unsigned         input   0b0111u          n bits   4u           bits    unsigned  malloc sizeof unsigned    n bits           bit   0       for bit   0  bit  lt  n bits    bit          bits bit     input  gt  gt  bit   amp  1       for bit   n bits  bit--           printf   u   bits bit        printf   n         free bits       Assuming that you want to calculate all bits as in this case  and not a specific one  the loop can be further changed to  for bit   0  bit  lt  n bits    bit  input  gt  gt   1      bits bit    input  amp  1    This modifies input in place and thereby allows the use of a constant width  single-bit shift  which may be more efficient on some architectures

User · Answer

If you don t want any loops  you ll have to write it out    include  lt stdio h gt   include  lt stdbool h gt   int main void        int num   7        if 0         bool arr 4       num amp 1   true  false   num amp 2   true  false   num amp 4   true  false   num amp 8   true  false         else          define BTB v i    v   amp   1u  lt  lt   i      true   false         bool arr 4      BTB num 0   BTB num 1   BTB num 2   BTB num 3             undef BTB      endif      printf   d  d  d  d n   arr 3   arr 2   arr 1   arr 0         return 0      As demonstrated here  this also works in an initializer

User · Answer

Here s one way to do it   there are many others   bool b 4   int v   7      number to dissect  for  int j   0   j  lt  4     j     b  j     0     v  amp   1  lt  lt  j      It is hard to understand why use of a loop is not desired  but it is easy enough to unroll the loop   bool b 4   int v   7      number to dissect  b  0     0     v  amp   1  lt  lt  0    b  1     0     v  amp   1  lt  lt  1    b  2     0     v  amp   1  lt  lt  2    b  3     0     v  amp   1  lt  lt  3      Or evaluating constant expressions in the last four statements   b  0     0     v  amp  1   b  1     0     v  amp  2   b  2     0     v  amp  4   b  3     0     v  amp  8

User · Answer

prateek thank you for your help   I rewrote the function with comments for use in a program   Increase 8 for more bits  up to 32 for an integer      std  vector  lt bool gt  bits from int  int integer        discern which bits of PLC codes are true       std  vector  lt bool gt  bool bits          continously divide the integer by 2  if there is no remainder  the bit is 1  else it s 0     for  int i   0  i  lt  8  i                  bool bits push back  integer 2         remainder of dividing by 2         integer    2        integer equals itself divided by 2            return bool bits

[c] How do I get bit-by-bit data from an integer value in C?

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