Internal docs for Rust/callback interface calls #2550
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| Calling into foreign code has 2 major differences from [calling Rust code](./rust_calls.md): | ||
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| * Foreign code is only called via an interface. | ||
| * The Rust code does not have access to a library with functions it can call. |
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I would like to reword this slightly. Like what does it mean? Does it mean that "Rust doesn't know the functions it can call on the foreign language side".
It clashes a bit with the next paragraph, which says that the foreign code registers functions Rust can call?
I think I know what you mean, but I had to read it a few times to get it.
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| VTable methods work similar to a [Rust method call](./rust_calls.md)], with some exceptions. | ||
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| Instead of returning the value as normal, it's written to an out pointer. |
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Super nit, but is it that returning means the foreign side writes the return value in a block of memory (heap or stack) and returns a pointer which points to the place in memory the Rust function can read the value from? Which then, it needs to "lift" and make it a Rust type?
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Not exactly, Rust passes in the pointer rather than having the foreign side return it. That makes the memory management a bit simpler. I updated the wording, tell me if it's clearer now.
| VTable methods work similar to a [Rust method call](./rust_calls.md)], with some exceptions. | ||
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| Instead of returning the value as normal, it's written to an out pointer. | ||
| This is because some foreign languages, like Python, have issues returning C structs like `RustBuffer`. |
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What does this mean, having issues returning C structs?
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It means ctypes throws an exception rather than returns the value. The design choice completely baffles me. I updated the text and linked to the issue for that.
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