6 ms·

It's not the same. With integer division you know the division rounds down, so using <= always splits between the two points. That said, it's always safer to c

• ced 10y ago
  ... unless max is a huge positive integer, and min is a huge negative integer!
  • RUBwkVjwLsDKgPw 10y ago
    Just use signed ints. Signed integer underflow is undefined behavior so it can't happen /s
    • gcc_programmer 10y ago
      Lol'd . To the contrary, it _can_ happen, and when it does the behaviour is undefined for C/C++. However, rest assured - you are probably on a x86_64 machine programming in sth like Ruby or Java script, so your Apps should be alright :P
• TimonKnigge 10y ago
  I like this one better: (min&max) + ((min^max) >> 1) #include <iostream> using namespace std; int avg(int a, int b) { return (a&b) + ((a^b)>>1); } int main() { cout << avg(int(2e9), int(2e9 + 10)) << endl; cout << avg(int(2e9), int(-2e9)) << endl; cout << avg(int(-2e9), int(-2e9 - 10)) << endl; return 0; } Gives: 2000000005 0 -2000000005
  • Retr0spectrum 10y ago
    Is there an explanation of how that actually works somewhere?
    • TimonKnigge 10y ago
      Yes there is! Right here: It's actually really simple. We'll write a and b in binary notation, for example: a = 1001101 b = 0100111 Now what happens when you add two numbers in binary? We essentially add the numbers in each column together, and if it overflows, we carry to the next column (this is how you carry out addition in general). So what are the columns where we need to carry, the ones that overflow? These are given by (a&b) - the columns where both a and b contain a one. To actually carry we just move everything one position to the left: ((a&b)<<1). And what are the columns where we don't need to carry, the ones that don't overflow? These are the ones where we have exactly one zero, either in a or in b, so: a^b. In other words, a + b = ((a&b)<<1) + (a^b). To compute the average, we divide by two, or in other words, we bitshift to the right by one place: (a + b)/2 = (a&b) + ((a^b)>>1) If anything is unclear, feel free to ask :)
      • Retr0spectrum 10y ago
        Great explanation, thanks.
    • pbsd 10y ago
      There's this well-known identity [1] a + b = (a ^ b) + ((a & b) << 1). This is essentially how a parallel ripple carry adder works. So (a + b) / 2 = (a + b) >> 1 = ((a ^ b) >> 1) + (a & b). [1] http://www.inwap.com/pdp10/hbaker/hakmem/boolean.html#item23 http://www.inwap.com/pdp10/hbaker/hakmem/boolean.html#item23
    • Avshalom 10y ago
      It's pretty obvious visually, you take the exclusive disjunct (a^b -> yellow) divide by two (>>1 -> bright yellow) and then add the intersection (a&b -> green) you use the full intersection because in the basic method (a+b / 2) they would each add their own half of the intersection. https://imgur.com/PUT4G9Z https://imgur.com/PUT4G9Z
  • luc4sdreyer 10y ago
    Reminds me of the way Java's binary search computes the average: int mid = (low + high) >>> 1; This was a fix because the original code had a bug: int mid =(low + high) / 2; Which was part of the Java code base for many years. Joshua Bloch based the code on Programming Pearls, which also contained the bug. More info: https://research.googleblog.com/2006/06/extra-extra-read-all-about-it-nearly.html https://research.googleblog.com/2006/06/extra-extra-read-all...
• jameshart 10y ago
  That is the safe technique for common cases such as locating a pivot in an array - where min and max are guaranteed to be positive (or you're using unsigned ints in the first place). That approach doesn't solve the overflow problem in general for signed ints.
• caf 10y ago
  Integer division doesn't round down, it rounds towards zero.