call-by-value

Concerning objects (especially strings), call by reference is faster than call-by-value because the function call does not need to create a copy of the original object. Using const, one can also ensure that the reference is not abused. My question is whether const call-by-reference is also faster if using primitive types, like bool, int or double. void doSomething(const string & strInput, unsigned int iMode); void doSomething(const string & strInput, const unsigned int & iMode); My suspicion is that it is advantageous to use call-by-reference as soon as the primitive type's size in bytes

This question already has an answer here: Is Java “pass-by-reference” or “pass-by-value”? 83 answers public class StackOverFlow { public static void main(String[] args) { ArrayList<String> al = new ArrayList<String>(); al.add("A"); al.add("B"); markAsNull(al); System.out.println("ArrayList elements are "+al); String str = "Hello"; markStringAsNull(str); System.out.println("str "+ str); } private static void markAsNull(ArrayList<String> str){ str.add("C"); str= null; } private static void markStringAsNull(String str){ str = str + "Append me"; str = null; } } This outputs: ArrayList elements are

Hello I am a beginner in C programming language. Recently I read about call by value and call by address. I have learned that in call by address changes in the called functions reflects the callee. However the following code does not work like that. int x = 10,y = 20; void change_by_add(int *ptr) { ptr = &y; printf("\n Inside change_by_add\t %d",*ptr); // here *ptr is printing 20 } void main(){ int *p; p = &x; change_by_add(p); printf("\nInside main\t %d", *p); // here *p is still pointing to address of x and printing 10 } When I am passing address then why the changes made by called function