pass-by-value

I got the answer NO! Because passing by value and passing by reference looks identical to the caller. However, the code below compiles right class A { public: void f(int i) {} void f(int& i) {} }; But when I try to use it, there is compile error. int main () { A a; int i = 9; int& j = i; a.f(1); a.f(i); a.f(j); return 0; } Why does not the compiler disable it even without knowing it is going to be used? Yes, they can be overloaded based on reference or not. That is why it's perfectly fine to have them coexist like that; they are different. The problem has to do with ambiguity. While f(1) can

This question already has an answer here: Is Java “pass-by-reference” or “pass-by-value”? 84 answers I was kind of baffled when I saw the following code did not work as expected. I thought Java always passed variables by references into functions. Therefore, why can't the function reassign the variable? public static void main(String[] args) { String nullTest = null; setNotNull(nullTest); System.out.println(nullTest); } private static void setNotNull(String s) { s = "not null!"; } This program outputs null . References to objects are passed by value in Java so assigning to the local variable

I have a general question about deep- and shallow-copy in the context of the pass-by-reference- and pass-by-value-concept of C#: In C# it is a requirement to explicitly create methods that accept pointers/references to be able to pass such to the method. However, at least objects passed as parameters to methods/constructors are behaving differently from the rest. It seems they are always passed by reference if no extra cloning is done as described here: http://zetcode.com/lang/csharp/oopii/ . Why are objects automatically passed by reference? Is there any particular benefit from forcing the

I'm fairly new to Java (been writing other stuff for many years) and unless I'm missing something (and I'm happy to be wrong here) the following is a fatal flaw... String foo = new String(); thisDoesntWork(foo); System.out.println(foo);//this prints nothing public static void thisDoesntWork(String foo){ foo = "howdy"; } Now, I'm well aware of the (fairly poorly worded) concept that in java everything is passed by "value" and not "reference", but String is an object and has all sorts of bells and whistles, so, one would expect that unlike an int a user would be able to operate on the thing that