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codesnik | on: Python 3.15's JIT is now back on track
basically all object oriented languages work like that. You access a member; you call a method which changes that member; you expect that change is visible lower in the code, and there're no statically computable guarantees that particular member is not touched in the called method (which is potentially shadowed in a subclass). It's not dynamism, even c++ works the same, it's an inherent tax on OOP. All you can do is try to minimize cost of that additional dereference. I'm not even touching threads here.

now, functional languages don't have this problem at all.

cherryteastain | parent
OOP has nothing to do with it. In your C++ example, foo(bar const&); is basically the same as bar.foo();. At the end of the day, whether passing it in as an argument or accessing this via the method call syntax it's just a pointer to a struct. Not to mention, a C++ compiler can, and often does, choose to put even references to member variables in registers and access them that way within the method call.

This is a Python specific problem caused by everything being boxed by default and the interpreter does not even know what's in the box until it dereferences it, which is a problem that extends to the "self" object. In contrast in C++ the compiler knows everything there's to know about the type of this which avoids the issue.

adrian17 | root | parent | next
That's not true. I mean: it's true that it has little to do with OOP, but most imperative languages (only exception I know is Rust) have the issue, it's not "Python specific". For example (https://godbolt.org/z/aobz9q7Y9):

struct S { const int x; int f() const; }; int S::f() const { int a = x; printf("hello\n"); int b = x; return a-b; }

The compiler can't reuse 'x' unless it's able to prove that it definitely couldn't have changed during the `printf()` call - and it's unable to prove it. The member is loaded twice. C++ compilers can usually only prove it for trivial code with completely inlined functions that doesn't mutate any external state, or mutates in a definitely-not-aliasing way (strict aliasing). (and the `const` don't do any difference here at all)

In Python the difference is that it can basically never prove it at all.

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