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> No where in here is there a requirement of sandboxing Let me quote the Webassembly spec [1]: > WebAssembly provides no ambient access to the computing envir

> No where in here is there a requirement of sandboxing Let me quote the Webassembly spec [1]: > WebAssembly provides no ambient access to the computing environment in which code is executed. Any interaction with the environment, such as I/O, access to resources, or operating system calls, can only be performed by invoking functions provided by the embedder and imported into a WebAssembly module. [1] https://webassembly.github.io/spec/core/intro/introduction.html#security-considerations https://webassembly.github.io/spec/core/intro/introduction.h...

devwastaken 7y ago
There is a seperation between spec and implimentation. That quote doesn't at all disagree with me. That's how wasm is meant to be, but doesn't give any detail as to how that's achieved programatically. You can impliment all of that and still have a vulnerable runtime because of how you implimented it. You can also just put in system access as functions, as some runtimes do because they don't care about sandboxing.
- thayne 7y ago
  > That's how wasm is meant to be, but doesn't give any detail as to how that's achieved programatically True, but a conforming implementation has at least some sandboxing, since it prevents arbitrary memory access. But the degree to which it is sandboxed depends on the functions that are exposed by the runtime.
- Dylan16807 7y ago
  > You can impliment all of that and still have a vulnerable runtime because of how you implimented it. This applies to every single sandboxed language in the world. The instructions are designed so that sandboxing the actual core is trivial, and the spec says that you have to sandbox outside of deliberate pass-throughs. That's about the best you can possibly do. If languages can qualify as sandboxed, it sounds like WASM qualifies. (And if they can't, then we're using a broken definition of "sandboxed".)
  - devwastaken 7y ago
    Languages cannot qualify as sandboxed unless that language comes with a runtime that has a sandbox. Wasm does not. There is no such thing as 'trivial to sandbox'. Wasm doesn't introduce anything new in terms of it's instructions, it's bound by all the same mistakes and errors developers will make in sandboxing as there has been in the past.
    - Dylan16807 7y ago
      So even if all possible runtimes have to be sandboxed or they're not actually implementing the language, it's not possible for a language to qualify as "sandboxed"? Then I stand by what I said before. Your definition is broken, and you're making a semantic argument rather than actually discussing eBPF and WASM. When you see someone say "sandboxed language" read it as "language where conforming implementations are by definition sandboxed". WASM meets that definition, as far as I can tell. When you see someone say "WASM is sandboxed" read it as "any runtime that implements the WASM spec is sandboxed".
      - devwastaken 7y ago
        If wasm had a standard runtime everyone used, sure. It doesn't, runtimes are significantly fragmented. Therefore 'wasm is sandboxed' is not true, and in many cases those sandboxes are not at all being audited. It is very dangerous to make broad and demonstrably untrue statements about software security. Wasm is not a sandbox, wasm is a set of instructions. Your quality of sandbox, if at all, is up to what runtime you use. No amount of word play will change that.
        Dylan16807 7y ago
        The defined semantics for those instructions include sandboxing. If there is no sandbox, it's not WASM. It wouldn't be implementing the instructions as described in the spec. You can argue that a sandbox might be low quality. That's fine. But it doesn't make it non-sandboxed.
wahern 7y ago
Those limitations are true of most virtual machines. Lua bytecode has no opcodes for doing I/O, invoking syscalls, or anything else that interacts outside the VM environment. AFAICT, Python is similar with the exception of some opcodes for printing to stdout. To interact with the outside environment you must load and invoke code from modules, so access is intrinsically limited by whatever modules are permitted to be loaded. In Lua a new VM state context has no modules loaded at all--not even for the string module; nor any ability to load modules--the C application needs to explicitly load the package module to register "require" in the environment.

devwastaken 7y ago

There is a seperation between spec and implimentation. That quote doesn't at all disagree with me. That's how wasm is meant to be, but doesn't give any detail as to how that's achieved programatically. You can impliment all of that and still have a vulnerable runtime because of how you implimented it. You can also just put in system access as functions, as some runtimes do because they don't care about sandboxing.

thayne 7y ago
> That's how wasm is meant to be, but doesn't give any detail as to how that's achieved programatically True, but a conforming implementation has at least some sandboxing, since it prevents arbitrary memory access. But the degree to which it is sandboxed depends on the functions that are exposed by the runtime.
Dylan16807 7y ago
> You can impliment all of that and still have a vulnerable runtime because of how you implimented it. This applies to every single sandboxed language in the world. The instructions are designed so that sandboxing the actual core is trivial, and the spec says that you have to sandbox outside of deliberate pass-throughs. That's about the best you can possibly do. If languages can qualify as sandboxed, it sounds like WASM qualifies. (And if they can't, then we're using a broken definition of "sandboxed".)
- devwastaken 7y ago
  Languages cannot qualify as sandboxed unless that language comes with a runtime that has a sandbox. Wasm does not. There is no such thing as 'trivial to sandbox'. Wasm doesn't introduce anything new in terms of it's instructions, it's bound by all the same mistakes and errors developers will make in sandboxing as there has been in the past.
  - Dylan16807 7y ago
    So even if all possible runtimes have to be sandboxed or they're not actually implementing the language, it's not possible for a language to qualify as "sandboxed"? Then I stand by what I said before. Your definition is broken, and you're making a semantic argument rather than actually discussing eBPF and WASM. When you see someone say "sandboxed language" read it as "language where conforming implementations are by definition sandboxed". WASM meets that definition, as far as I can tell. When you see someone say "WASM is sandboxed" read it as "any runtime that implements the WASM spec is sandboxed".
    - devwastaken 7y ago
      If wasm had a standard runtime everyone used, sure. It doesn't, runtimes are significantly fragmented. Therefore 'wasm is sandboxed' is not true, and in many cases those sandboxes are not at all being audited. It is very dangerous to make broad and demonstrably untrue statements about software security. Wasm is not a sandbox, wasm is a set of instructions. Your quality of sandbox, if at all, is up to what runtime you use. No amount of word play will change that.
      - Dylan16807 7y ago
        The defined semantics for those instructions include sandboxing. If there is no sandbox, it's not WASM. It wouldn't be implementing the instructions as described in the spec. You can argue that a sandbox might be low quality. That's fine. But it doesn't make it non-sandboxed.

wahern 7y ago

Those limitations are true of most virtual machines. Lua bytecode has no opcodes for doing I/O, invoking syscalls, or anything else that interacts outside the VM environment. AFAICT, Python is similar with the exception of some opcodes for printing to stdout. To interact with the outside environment you must load and invoke code from modules, so access is intrinsically limited by whatever modules are permitted to be loaded. In Lua a new VM state context has no modules loaded at all--not even for the string module; nor any ability to load modules--the C application needs to explicitly load the package module to register "require" in the environment.