Even if you had a way to pass the array list as a copy and not by reference it would have been only a shallow copy. I would do something like: void foo(final ArrayList list) { ArrayList listCopy = new ArrayList(list); // Rest of the code } And just work on the copied list.
In JavaScript objects are always passed by copy-reference. I’m not sure if that’s the exactly correct terminology, but a copy of the reference to the object will be passed in. This means that any changes made to the object will be visible to you after the function is done executing. Code: var obj = { … Read more
Yes, use a closure: functionName($someArgument, function() use(&$variable) { $variable = “something”; }); Note that in order for you to be able to modify $variable and retrieve the modified value outside of the scope of the anonymous function, it must be referenced in the closure using &.
I think much confusion is generated by not communicating what is meant by passed by reference. When some people say pass by reference they usually mean not the argument itself, but rather the object being referenced. Some other say that pass by reference means that the object can’t be changed in the callee. Example: struct … Read more
In general, using a non const reference like Ref<T>& is never a good idea because this will only work if the caller already has a Ref<T> object at hand. This discards 5 and 7. The cases 3 and 4 does not make sense, and they won’t compile in general. Then, there is not much difference … Read more
There is one thing nobody mentioned. There is a certain GCC optimization called IPA SRA, that replaces “pass by reference” with “pass by value” automatically: https://gcc.gnu.org/onlinedocs/gcc/Optimize-Options.html (-fipa-sra) This is most likely done for scalar types (eg. int, double, etc), that does not have non-default copy semantics and can fit into cpu registers. This makes void(const … Read more
To assign a reference in a constructor you need to have a reference member class A{ std::string& str; public: A(std::string& str_) : str(str_) {} }; str is now a reference to the value you passed in. Same applies for const refs class A{ const std::string& str; public: A(const std::string& str_) : str(str_) {} }; However … Read more
Java always passes arguments by value, NOT by reference. In your example, you are still passing obj by its value, not the reference itself. Inside your method changeName, you are assigning another (local) reference, obj, to the same object you passed it as an argument. Once you modify that reference, you are modifying the original … Read more
A dictionary is a reference type, so it is not possible to pass by value, although references to a dictionary are values. Let me try to clear this up: void Method1(Dictionary<string, string> dict) { dict[“a”] = “b”; dict = new Dictionary<string, string>(); } void Method2(ref Dictionary<string, string> dict) { dict[“e”] = “f”; dict = new … Read more