bit-shift

I recently found this piece of JavaScript code: Math.random() * 0x1000000 << 0 I understood that the first part was just generating a random number between 0 and 0x1000000 (== 16777216). But the second part seemed odd. What's the point of performing a bit-shift by 0? I didn't think that it would do anything. Upon further investigation, however, I noticed that the shift by 0 seemed to truncate the decimal part of the number. Furthermore, it didn't matter if it was a right shift, or a left shift, or even an unsigned right shift. > 10.12345 << 0 10 > 10.12345 >> 0 10 > 10.12345 >>> 0 10 I tested

In all versions of C and C++ prior to 2014, writing 1 << (CHAR_BIT * sizeof(int) - 1) caused undefined behaviour, because left-shifting is defined as being equivalent to successive multiplication by 2 , and this shift causes signed integer overflow: The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. [...] If E1 has a signed type and nonnegative value, and E1 × 2 E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined . However in C++14 the text has changed for << but not for multiplication: The

Possible Duplicate: Why is (-1 >>> 32) = -1? The unsigned right shift operator inserts a 0 in the leftmost. So when I do System.out.println(Integer.toBinaryString(-1>>>30)) output 11 Hence, it is inserting 0 in the left most bit. System.out.println(Integer.toBinaryString(-1>>>32)) output 11111111111111111111111111111111 Shouldn't it be 0? See http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.19 If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were

What method would you use to determine if the the bit that represents 2^x is a 1 or 0 ? I'd use: if ((value & (1L << x)) != 0) { // The bit was set } (You may be able to get away with fewer brackets, but I never remember the precedence of bitwise operations.) Another alternative: if (BigInteger.valueOf(value).testBit(x)) { // ... } Ande I wonder if: if (((value >>> x) & 1) != 0) { } .. is better because it doesn't matter whether value is long or not, or if its worse because it's less obvious. Tom Hawtin - tackline Jul 7 at 14:16 You can also use bool isSet = ((value>>x) & 1) != 0; EDIT: the