[C] Inverts the binary of a decimal number and outputs it

describe

  Here are three ways ( The third kind of accidental thought , Ha ha ha , Very simple )

  The first way
     This solution is more conventional , The idea is to initialize an array ( The default value is 0),
Then take each one out , Replace the corresponding value in the array , Because when you take binary
It is taken from right to left , And the array is stored from left to right , So the number you get
A group is a sequence in reverse order .

void reverse_bits(unsigned int value)
{
	int result_buffer[32] = { 0 };
	int* Tag = result_buffer;
	while (value != 0)
	{
		int temp = value / 2;
		*Tag++ = value - temp * 2;
		value /= 2;
	}
	int i;
	for (i = 0; i < 32; i++)
	{
		printf("%d", result_buffer[i]);
	}
}

The second way

    This is an improved version of the previous method , In a unique way ,
The binary number is used to verify the original number ( namely 2 To the power of , This is used for verification
In binary form, only one bit is 1), Make this decimal by logical and
Count it down , And stored in the array .

void reverse_bits(unsigned int value)
{
	int result_buffer[32] = { 0 };
	int i;
	for (i = 0; i < 32; i++)
	{
		int temp = (value & (int)pow(2,i)) != 0 ? 1 : 0;
		result_buffer[i] = temp;
	}

	for (i = 0; i < 32; i++)
	{
		printf("%d", result_buffer[i]);
	}
}

The third way

    This method is based on the second method , Since we can

Find each bit , So why can't we just put together this number in binary ?

Simply move left , Relative to the limitations of getting an array , A concrete

The values are obviously more flexible , And more in line with the meaning of the topic .

unsigned int reverse_bits(unsigned int value)
{
	unsigned int result = 0;
	int i;
	for (i = 0; i < 32; i++)
	{
		result += (((value & (int)pow(2, i)) != 0 ? 1 : 0) << (31-i));
	}
	printf("%u", result);
	return result;
}