Clone a struct without copying the memory address

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Clone a struct without copying the memory address

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#1 by (1 votes)

1

I need to copy struct problema1 to struct problema2 , but when doing the way you are in the program below, when I change the struct problema2 , you are also changing the struct problema1 . The way I did it by copying the reference, I wanted to copy without reference, but I do not know how to do it.

void __fastcall TForm1::Button1Click(TObject *Sender)
{

 struct Tproblem
   {
    TList *Lista;
    float x;
    };

  typedef struct Tproblem Problem;
  Problem *Problema1 ,*Problema2;
  int *x;
  Problema1 = new Problem;
  Problema1->Lista = new TList;
  for( int i = 0 ; i<10; i++)
    {
       x = new int ;
       *x = i;
       Problema1->Lista->Add(x) ;
     }



   Problema2 = new Problem;
    Problema2 = Problema1;// aki eu copia a estrutura


     for( int i = 0 ; i<Problema1->Lista->Count ;i++)
    {
      x = (int*)Problema1->Lista->Items[i];
      ListBox1->Items->Strings[i]=IntToStr(*x);

c++ struct c++-builder

asked by anonymous 08.07.2016 / 18:08

1 answer

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score 1 · Accepted Answer

This code is copying the pointer, so the code has two variables pointing to the same object, so changing one changes the other. What you need to do is create a new object and point to it. This can be done with memcpy() . Something like this:

memcpy(Problema2, Problema1, sizeof(Problema1));

There are better ways to solve this, but then you would need to tinker with the whole code.