3 ms·

I don't understand. What does the "single assignment" bit get you that a compiler with an SSA pass doesn't?

jldugger 7y ago
I assume the removal of pointer aliasing?
gengkev 7y ago
Probably that the original binding of p is immutable, as it would be in a functional language: you can create a new binding, but you can't actually change the value of the existing binding.
- nwallin 7y ago
  Unless there's something I don't understand, that's what the single static assignment pass does in a compiler.
  - AgentME 7y ago
    New bindings live at different addresses (if you pass a pointer to one binding, and then make a new binding with a new value, the existing pointer will point to the old value and not the new binding's value) and bindings made in a loop iteration won't live to the next iteration.
    - lostmsu 7y ago
      This sounds like an implementation detail. What does the language spec say about the outcome.
      - AgentME 7y ago
        Different bindings are like entirely different variables that just coincidentally share the same name.
      - a1369209993 7y ago
        The (C) language spec says that: x = 5; int* p = &x; x = 6; // new binding of x printf("%i\n",*p); prints "6". If new bindings lived at different addresses, it would print "5".
  - coldtea 7y ago
    When the SSA pass happens in the compiler it's already too late to have guaranteed certain behaviours from the source stage... E.g. you can turn a C program into SSA, but if you have aliasing issues, it's not gonna fix them...
dragonwriter 7y ago
An SSA pass is an implementation detail that can't change language semantics, so, as I understand, it either doesn't transform or applies only to pointers when values have externally (to the block in which they are defined) visible mutation based on the language semantics; having single assignment semantics and thereby lacking mutability is a significant property for, particularly, parallel code with shared references. If no one can mutate a reference, you don't need Rust- or Pony-like tracking of who can see or mutate it, it's safe for everyone to see it.

I assume the removal of pointer aliasing?

Probably that the original binding of p is immutable, as it would be in a functional language: you can create a new binding, but you can't actually change the value of the existing binding.

nwallin 7y ago
Unless there's something I don't understand, that's what the single static assignment pass does in a compiler.
- AgentME 7y ago
  New bindings live at different addresses (if you pass a pointer to one binding, and then make a new binding with a new value, the existing pointer will point to the old value and not the new binding's value) and bindings made in a loop iteration won't live to the next iteration.
  - lostmsu 7y ago
    This sounds like an implementation detail. What does the language spec say about the outcome.
    - AgentME 7y ago
      Different bindings are like entirely different variables that just coincidentally share the same name.
    - a1369209993 7y ago
      The (C) language spec says that: x = 5; int* p = &x; x = 6; // new binding of x printf("%i\n",*p); prints "6". If new bindings lived at different addresses, it would print "5".
- coldtea 7y ago
  When the SSA pass happens in the compiler it's already too late to have guaranteed certain behaviours from the source stage... E.g. you can turn a C program into SSA, but if you have aliasing issues, it's not gonna fix them...

dragonwriter 7y ago

An SSA pass is an implementation detail that can't change language semantics, so, as I understand, it either doesn't transform or applies only to pointers when values have externally (to the block in which they are defined) visible mutation based on the language semantics; having single assignment semantics and thereby lacking mutability is a significant property for, particularly, parallel code with shared references. If no one can mutate a reference, you don't need Rust- or Pony-like tracking of who can see or mutate it, it's safe for everyone to see it.