Java is Pass-by-Value, Dammit

Well the way I've always thought of it is that Java passes "references by value". Meaning the target method receives a _copy_ of the pointer, not the actual pointer. So the result of newing up the parameter inside the target method doesn't affect the original, because you're just making the pointer point to a NEW memory address. On the other hand, mutating the parameter WILL affect the original object, because a copy of a pointer still points to that object…

This article is overly pedantic. Java's semantics are pass-by-value only of you consider that the "values" that are being passed are pointers. But Java doesn't have first-class pointers, it has implicit pointers. The only thing you can do with a Java pointer "value" is to tacitly dereference it. In fact, you can write a swap(x,y) function in Java. It goes something like this: class foo { int value } void swap(foo x, foo y} { int temp = x.value; x.value…

Good post.... I am a Java developer with almost 10 yrs of experience and I know I often make that mistake even on Job interviews:).... The problem is that people used to Java tend to think more about the intuitive consequence of passing by ref vs value. When a Java dev thinks of passing by reference he's thinking: "I can alter the state of the reference and the changes would be visible globally, including from within the calling context". In…

If "Object references are passed by value", then the "Object" itself is passed by reference, so the statement "In Java, Objects are passed by reference" it is a full true statement. You can pass "Object references" by value or by reference, anyway, since you're passing a "Object reference", the object itself is being passed as a reference.