In Java, are parameters accesses in order when passed into a method?

This is important because I am creating an object out of deserialized primitives within the readObject method. So if I am writing

ObjectInputStream s; //From the readObject method
Object obj = new Object(s.readDouble(), s.readDouble开发者_运维百科(), s.readDouble());

Will it deserialize in the correct order, the order they were written to the byte stream?

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Yes parameters will be evaluated in order.

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In Java, expressions are always evaluated left-to-right. So the result of the first readDouble call will become the first argument to the Object constructor (but please don't name your class Object! that collides with java.lang.Object), and so on.

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Yes, they will deserialize in the correct order.

You might want to create (good) named variables out of those s.readDouble() values though, it's more clear what they mean.

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Java argument lists are evaluated Left-to-Right