将字节转换为位数组？ C

Question

I want to read binary file byte at the time and then store bits of that byte into integer array. And similarly I want to write integer array of 1s and 0s (8 of them ) into binary file as bytes?

Answer 1

If you have an array of bytes:

unsigned char bytes[10];

And want to change it into an array of bits:

unsigned char bits[80];

And assuming you have 8 bits per byte, try this:

int i;
for (i=0; i<sizeof(bytes)*8; i++) {
    bits[i] = ((1 << (i % 8)) & (bytes[i/8])) >> (i % 8);
}

In this loop, i loops through the total number of bits. The byte that a given bit lives at is i/8 , which as integer division rounds down. The position of the bit within a byte is i%8 .

First we create a mask for the desired bit:

1 << (i % 8)

Then the desired byte:

bytes[i/8]

Then we perform a logical AND to clear all bits except the one we want.

(1 << (i % 8)) & (bytes[i/8])

Then we shift the result right by the bit position to put the desired bit at the least significant bit. This gives us a value of 1 or 0.

Note also that the arrays in question are unsigned . That is required for the bit shifting to work properly.

To switch back:

int i;
memset(bytes, 0, sizeof(bytes));
for (i=0; i<sizeof(bytes)*8; i++) {
    bytes[i/8] |= bits[i] << (i % 8);
}

We start by clearing out the byte array, since we'll be setting each byte one bit at a time.

Then we take the bit in question:

bits[i]

Shift it into its position:

bits[i] << (i % 8)

Then use a logical OR to set the appropriate byte;

Answer 2

A simple C program to do the job on a byte array 'input' of size 'sz' would be:

int i=0,j=0;
unsigned char mask = 0x01u;    

for (i=0;i<sz;i++)
        for (j=0;j<8;j++)
           output[8*i+j]=((unsigned char)input[i] >> j) & (unsigned char)(mask);