I'm not sure what you mean by "same behavior". I don't think you can externally distinguish between "x = x + x" and "x += x" in Rust, because while in the first case a new temporary value will be created, it will overwrite the left side value: x will be the "same object" as it's in the same address in both cases. In Python objects are stored indirectly, so after an assignment x will then point to a new object instead of the old one.
But yes, not making copies is more efficient so the same optimization applies.
Nevertheless, in Rust (or C++) the parameter x would not have been passed as a mutable reference into the function, so the value provided as input would not have been mutated.
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I'm not sure what you mean by "same behavior". I don't think you can externally distinguish between "x = x + x" and "x += x" in Rust, because while in the first case a new temporary value will be created, it will overwrite the left side value: x will be the "same object" as it's in the same address in both cases. In Python objects are stored indirectly, so after an assignment x will then point to a new object instead of the old one.
But yes, not making copies is more efficient so the same optimization applies.
Nevertheless, in Rust (or C++) the parameter x would not have been passed as a mutable reference into the function, so the value provided as input would not have been mutated.