Python 3.6 has added Flag and IntFlag which support the usual bit-wise operations. As a bonus, the resulting values from the bit-wise operations are still members of the original flag class, and are singletons [1]. The aenum library also has this addition and is usable back to Python 2.7. [1] A bug exists in 3.6.0: … Read more
When i gets to -1, the value of ~i is ~-1, or 0, so the while loop stops executing. The ! operator works because it does something completely different; it results in 1 for 0 values and 0 for all other values. ~ is a bitwise negation. A little more in detail: ~ takes each … Read more
It’s not a bitwise operator when it’s applied to boolean operators. It’s the same as: someBoolean = someBoolean & someString.ToUpperInvariant().Equals(“blah”); You usually see the short-cut and operator &&, but the operator & is also an and operator when applied to booleans, only it doesn’t do the short-cut bit. You can use the && operator instead … Read more
The number 0177 is an octal number representing the binary pattern below: 0000000001111111 When you AND it using the bitwise operation &, the result keeps the bits of the original only in the bits that are set to 1 in the “mask”; all other bits become zero. This is because “AND” follows this rule: X … Read more
In short, (high + low) >>> 1 is a trick that uses the unused sign-bit to perform a correct average of non-negative numbers. Under the assumption that high and low are both non-negative, we know for sure that the upper-most bit (the sign-bit) is zero. So both high and low are in fact 31-bit integers. … Read more
There are 2 integers between 1 and -2: 0 and -1 1 in binary is 00000000000000000000000000000001 0 in binary is 00000000000000000000000000000000 -1 in binary is 11111111111111111111111111111111 -2 in binary is 11111111111111111111111111111110 (“binary” being 2’s complement, in the case of a bitwise not ~ ) As you can see, it’s not very surprising ~1 equals -2, … Read more
To set a bit, use: x |= 0b1; // set LSB bit x |= 0b10; // set 2nd bit from LSB to erase a bit use: x &= ~0b1; // unset LSB bit (if set) x &= ~0b10; // unset 2nd bit from LSB to toggle a bit use: x ^= 0b1; Notice I use … Read more
Compare the binary representations of each of those. 110 & 010 = 010 1010 & 0101 = 0000 10100 & 11001 = 10000 1111011 & 0010100 = 0010000 In each case, a digit is 1 in the result only when it is 1 on both the left AND right side of the input.
Cast the number to unsigned and then shift. That will force a 0-fill.
You use the modulo operator: while(score) { printf(“%d\n”, score % 10); score /= 10; } Note that this will give you the digits in reverse order (i.e. least significant digit first). If you want the most significant digit first, you’ll have to store the digits in an array, then read them out in reverse order.