Array has garbage values in function in c++!

Question

I have the following function:

  int *array1=new int [3];
  void My_function()
  {
  My_set(array1);
  for(int i=0;i<3;i++){    //The output is 5 15 55
  cout<<array1[i]<<endl;
  }
  Display (array1)

  }

  void My_set(int *array1)
  {
  array1[0]=5;
  array1[1]=15;
  array1[2]=55;
   }

  void Display(int *array1){
  for(int i=0;i<3;i++){
  cout<<array1[i]<<endl; //The output is Garbage  -842150451
   }

   }

Note: This problem occurred in complex project but I show my problem in simple code ! THANK YOU :)

Use `std::vector` and it will warn you when you overflow the buffer in most debug compilers. — Neil Kirk, Apr 12 '14 at 23:08
@awesomeyi: That code you compiled has undefined behavior. So it's irrelevant that it does what you expected it to do. — Benjamin Lindley, Apr 12 '14 at 23:15
Generally use of raw pointers is not a good idea. In your case where number of elements is already known use 'std::array' (e.g., 'std::array array1'). — 101010, Apr 13 '14 at 00:41

score 0 · Answer 1 · answered Apr 12 '14 at 23:36

0

You need to declare your array like this int *array1 = new int[3];

answered Apr 12 '14 at 23:36

Inertiatic

1,270
1
9
12

Definitely, I was wondering why he/she was declaring it with new. Figured for testing? Dan brought up a good point though. @user3316887 if you end up creating your array with a dynamic size, make sure to pass the size of the array as an argument to your functions. – Inertiatic Apr 13 '14 at 00:19

Array has garbage values in function in c++!

1 Answers1