bytecode | Python module to modify bytecode | Bytecode library

Had this issue myself, apparently it's some sort of Ganache CLI error but the simplest fix I could find was to change the network id in Ganache through settings>server to 1337. It restarts the session so you'd then need to change the address and private key variable.

If it's the same tutorial I'm doing, you're likely to come unstuck after this... the code for transaction should be:

You can not turn arbitrary classes into hidden classes.

The documentation of defineHiddenClass contains the sentence

• On any attempt to resolve the entry in the run-time constant pool indicated by this_class, the symbolic reference is considered to be resolved to C and resolution always succeeds immediately.

What it doesn’t spell out explicitly is that this is the only place where a type resolution ever ends up at the hidden class.

But it has been said unambiguously in bug report JDK-8222730:

For a hidden class, its specified hidden name should only be accessible through the hidden class's 'this_class' constant pool entry.

The class should not be accessible by specifying its original name in, for example, a method or field signature even within the hidden class.

Which we can check. Even a simple case like

1. Yes, CPython is compiled to bytecode which is then executed by the virtual machine.
2. The virtual machine executes instructions one-by-one. It's written in C (but you can write it in another language) and looks like a huge if/else statement like "if the current instruction is this, do this; if the instruction is this, do another thing", and so on. Instructions aren't translated to binary - that's why it's called an interpreter.
   1. You can find the list of instructions here: https://docs.python.org/3.10/library/dis.html#python-bytecode-instructions
   2. The implementation of the VM is available here: https://github.com/python/cpython/blob/f71a69aa9209cf67cc1060051b147d6afa379bba/Python/ceval.c#L1718
3. Bytecode doesn't have a concept of "line": it's just a stream of bytes. The interpreter can read one byte at a time and use another if/else statement to decide what instruction it's looking at. For example:

I have you tried to deploy with the argument --verbose?

That should be something like that (not sure of the syntax because I am on phone) erdpy --verbose contract deploy

V8 developer here.

are small integers pooled or not?

They are not (at least not right now). That said, this is a small implementation detail and could be done either way: it would totally be possible to use the constant pool for Smis. I suppose the decision to build special machinery for Smis (instead of reusing the general-purpose constant pool) was made because things turned out to be more efficient that way.

it is not worth it pooling small integers if sizeof(int) < sizeof(int *)

The details are different (a Smi is not an int, and constant pool slots are referenced by index rather than C++ pointer), but this reasoning does go in the right direction: avoiding indirections can save time and memory.

are smis also allocated on the heap?

Yes, everything is allocated on the heap. The stack is only useful for temporary (and sufficiently small) things; that's largely unrelated to the type of thing.

The "trick" of Smis is that they're not stored as separate objects: when you have an object that refers to a Smi, such as let foo = {smi: 42}, then the value 42 can be smi-encoded and stored directly inside the "foo" object (whereas if the value was 42.5, then the object would store a pointer to a separate "HeapNumber"). But since the object is on the heap, so is the Smi.

@DanielCruz

What I understand [...] is that constant small integers are pooled. Variable small integers are not.

Nope. Any literal that occurs in source code is "constant". Whether you use let or const for your variables has nothing to do with this.

The short answer to your question is, there is a difference between explicitly calling a function with an argument, compared to calling the function and letting it supply the same argument as default by itself. In the latter case, the caller environment is the execution environment of the function. In the former case the caller environment is where the function was explicitly called from.

It took me a while to find a nice visualization of what is going on, but I think I found one.

Let's create two functions similar to {rlang}'s env_parent and caller_env. The difference is that myCallEnv will not only return the execution environment of the calling function, it will also show the call strack tree with lobstr::cst():

TL;DR: You are observing what happens when JIT compiler cannot trust that values are not changing inside the loop. Additionally, in the tiny benchmark like this, Blackhole.consume costs dominate, obscuring the results.

Simplifying the test:

Instead of this:

It is a problem with your Eclipse, not with Java-17 itself. Java-17 has been released only yesterday. Wait for some time until the IDEs are updated to support Java-17.

Demo: