4 ms·

Wouldn’t a stack map have that value as well?

• kazinator 2y ago
  If the compiler puts out a stack map that is conservative, then the GC scan will be effectively conservative. The compiler has to compensate for that somehow. If a temporary location is in the stackmap, such that the value in it is a dead value before a function call, the compiler has to insert an instruction to null out that temporary location.
  • Rusky 2y ago
    And the same can of course be done under conservative GC...
    • kazinator 2y ago
      But it won't, because people implement conservative GCs when they are not in control of all the pieces, like the compilers used for some parts of their run-time.
      • Rusky 2y ago
        The point of the post is that conservative GC can be useful for reasons other than that simple lack of control.