• ZILtoid1991@lemmy.worldOP
      61·
      18 days ago

      The & operator references the value.

      int i;
int* p = &i;

      In C++, the & at the function argument makes it a reference type (safe pointer).

      void someFunction(int& refVal) {
    [...]
}
      • Flames5123@sh.itjust.worksEnglish
        32·
        18 days ago

        As someone who likes working with higher level languages, I never understood the pass by reference or even referencing different pointers. It never stuck out to me as useful in what I want software to do. It’s too close to hardware.

        • lime!@feddit.nu
          9·
          18 days ago

          if you’re working with higher level languages you pass-by-reference all the time. give a list to a function to modify it? that’s by reference. giving an event handler function to a framework? that’s by reference. setting a property on an object? that’s usually by reference.

          • funkless_eck@sh.itjust.works
            11·
            18 days ago

            the list is the helpful part to understanding it.

            it would be terrible if, with bar being a list and foo being a member of the list

             if foo in bar: return true

            modified the list. So yeah, you want to look at the list not edit the list, it’s a pointer.

            • lime!@feddit.nu
              3·
              18 days ago

              other way round surely? if you want to modify the original object, use a pointer. if you don’t, use a copy.

            • gerryflap@feddit.nl
              1·
              18 days ago

              Your example doesn’t make sense to me. Why would you modify the entire list by checking if something is in it? Also, you can totally edit the list via a pointer, that’s how you’re supposed to edit the list if you want any performance. Otherwise you’d be copying the list on every modification, which is terribly inefficient

        • ZILtoid1991@lemmy.worldOP
          1·
          18 days ago

          Reference values are quite useful, such as:

          double valOut;
if (parseDouble(valOut) == 0) { //Argument of parseDouble is ref type, no & needed for input, no exceptions needed for error handling
    [...] //No error, code executed normally
} else {
    [...] //Erroreus input
}
    • dejected_warp_core@lemmy.world
      3·
      18 days ago

      *x = dereference or “point to”. Treats the variable x as containing a pointer value. Evaluates to a variable existing at the address in x.

      &x = reference or “get address of”. Evaluates to the address of x.

      They’re complimentary operators, so *(&x) cancels out and is equvalent to just x.