That's just the placeholder implementation that will work on any platform and even with the interpreter.

If you look at the openjdk9 sources you will notice that it is annotated as intrinsic candidate[0]. But earlier versions also have intrinsics for that[1], it's just not annotated as such.

[0] http://hg.openjdk.java.net/jdk9/jdk9/jdk/file/23721aa1d87f/s... [1] https://gist.github.com/apangin/7a9b7062a4bd0cd41fcc#file-ho...

> obvious that it's not going to use the popcnt opcode instruction ... I'd guess totally impossible in Java

I don't understand why you would guess at what Java can do when you can find out for certain?

    public class Test {
      
      private static int bitCount(int a) {
        return Integer.bitCount(a);
      }
      
      public static void main(String[] args) {
        while (true) {
          bitCount(14);
        }
      }
      
    }

Compiles bitCount to

    0x000000011e637980: sub    rsp,0x18
    0x000000011e637987: mov    QWORD PTR [rsp+0x10],rbp  ;*synchronization entry
                                                  ; - Test::bitCount@-1 (line 4)

    0x000000011e63798c: popcnt eax,esi            ;*invokestatic bitCount
                                                  ; - Test::bitCount@1 (line 4)

    0x000000011e637990: add    rsp,0x10
    0x000000011e637994: pop    rbp
    0x000000011e637995: test   DWORD PTR [rip+0xfffffffff10ed665],eax        # 0x000000010f725000
                                                  ;   {poll_return}
    0x000000011e63799b: ret 

There's your popcnt instruction. No need to guess.

There are methods in Java which have special vm implementations to take advantage of instructions like this. They methods have an ordinary Java code version for the interpreter and on jit compilation are replaced with the other version. For a complete list of these types of functions see http://hg.openjdk.java.net/jdk8/jdk8/hotspot/file/87ee5ee275...

JIT-compiler might recognize this call and replace it with popcnt instruction. I'm not sure if it does that, though.