C: Building a byte | 易学教程

问题

I have an array with 16 elements. I would like to evaluate these to a boolean 0 or 1 and then store this in 2 bytes so i can write to a binary file. How do I do this?

回答1:

Something like this you mean?

unsigned short binary = 0, i;
for ( i = 0; i < 16; ++i )
  if ( array[i] )
    binary |= 1 << i; 

// the i-th bit of binary is 1 if array[i] is true and 0 otherwise.

回答2:

You have to use bitwise operators.

Here's an example:

int firstBit = 0x1;
int secondBit = 0x2;
int thirdBit = 0x4;
int fourthBit = 0x8;

int x = firstBit | fourthBit; /*both the 1st and 4th bit are set */
int isFirstBitSet = x & firstBit; /* Check if at least the first bit is set */

回答3:

int values[16];
int i;
unsigned short word = 0;
unsigned short bit = 1;

for (i = 0; i < 16; i++)
{
    if (values[i])
    {
        word |= bit;
    }

    bit <<= 1;
}

回答4:

This solution avoid the use of the if inside the loop:

unsigned short binary = 0, i;
for ( i = 0; i < 16; ++i )
  binary |= (array[i] != 0) << i;

回答5:

Declare an array result with two bytes, then you loop through the source array:

for (int i = 0; i < 16; i++) {
  // calclurate index in result array
  int index = i >> 3;
  // shift value in result
  result[index] <<= 1;
  // check array value
  if (theArray[i]) {
    // true, so set lowest bit in result byte
    result[index]++;
  }
}

回答6:

Something like this.

int values[16];
int bits = 0;
for (int ii = 0; ii < 16; ++ii)
{
    bits |= (!!values[ii]) << ii;
}

unsigned short output = (unsigned short)bits;

the expression (!!values[ii]) forces the value to be 0 or 1, if you know for sure that the values array already contains either a 0 or a 1 and nothing else, you can leave of the !!

You could also do this if you don't like the !! syntax.

int values[16];
int bits = 0;
for (int ii = 0; ii < 16; ++ii)
{
    bits |= (values[ii] != 0) << ii;
}

unsigned short output = (unsigned short)bits;

来源：https://stackoverflow.com/questions/2393914/c-building-a-byte

标签

binary

byte