Because the array is being passed by value, an exact copy of the array is made and placed on the stack. This is incorrect: the array itself is not being copied, only a copy of the pointer to its address is passed to the callee (placed on the stack). (Regardless of whether you declare the … Read more
Yes to both. Pointers are passed by value as anything else. That means the contents of the pointer variable (the address of the object pointed to) is copied. That means that if you change the value of the pointer in the function body, that change will not be reflected in the external pointer that will … Read more
You can change what foo points to using y: Foo foo = new Foo(“1”); void Bar(ref Foo y) { y = new Foo(“2”); } Bar(ref foo); // foo.Name == “2”
The reason for this bug is that the data used by the create_array function is a local variable that only exists inside that function. The assigned memory address obtained from malloc is only stored in this local variable and never returned to the caller. Consider this simple example: void func (int x) { x = … Read more
It’s interesting in JavaScript. Consider this example: function changeStuff(a, b, c) { a = a * 10; b.item = “changed”; c = {item: “changed”}; } var num = 10; var obj1 = {item: “unchanged”}; var obj2 = {item: “unchanged”}; changeStuff(num, obj1, obj2); console.log(num); console.log(obj1.item); console.log(obj2.item); This produces the output: 10 changed unchanged If obj1 was … Read more
Python passes references-to-objects by value. Python passes references-to-objects by value (like Java), and everything in Python is an object. This sounds simple, but then you will notice that some data types seem to exhibit pass-by-value characteristics, while others seem to act like pass-by-reference… what’s the deal? It is important to understand mutable and immutable objects. … Read more
Java always passes arguments by value NOT by reference. Let me explain this through an example: public class Main { public static void main(String[] args) { Foo f = new Foo(“f”); changeReference(f); // It won’t change the reference! modifyReference(f); // It will modify the object that the reference variable “f” refers to! } public static … Read more
It’s by value according to the PHP Documentation. By default, function arguments are passed by value (so that if the value of the argument within the function is changed, it does not get changed outside of the function). To allow a function to modify its arguments, they must be passed by reference. To have an … Read more
Stop a Variable’s Reassignment While these answers are intellectually interesting, I’ve not read the short simple answer: Use the keyword final when you want the compiler to prevent a variable from being re-assigned to a different object. Whether the variable is a static variable, member variable, local variable, or argument/parameter variable, the effect is entirely … Read more