reference to pointer to int causes error

Question

I have the following piece of code, which is an implementation of an array resizing function. It seems to be correct, but when I compile the program I get the following errors:

g++ -Wall -o "resizing_arrays" "resizing_arrays.cpp" (in directory: /home/aristofanis/Desktop/coursera-impl)
resizing_arrays.cpp: In function ‘int main()’:
resizing_arrays.cpp:37: error: invalid initialization of non-const reference of type ‘int*&’ from a temporary of type ‘int*’
resizing_arrays.cpp:7: error: in passing argument 1 of ‘void resize(int*&, int, int, int)’
resizing_arrays.cpp:39: error: invalid initialization of non-const reference of type ‘int*&’ from a temporary of type ‘int*’
resizing_arrays.cpp:7: error: in passing argument 1 of ‘void resize(int*&, int, int, int)’
resizing_arrays.cpp:41: error: invalid initialization of non-const reference of type ‘int*&’ from a temporary of type ‘int*’
resizing_arrays.cpp:7: error: in passing argument 1 of ‘void resize(int*&, int, int, int)’
Compilation failed.

Here is the code:

int N=5;

void resize( int *&arr, int N, int newCap, int initial=0 ) { // line 7
  N = newCap;
  int *tmp = new int[ newCap ];
  for( int i=0; i<N; ++i ) {
    tmp[ i ] = arr[ i ];
  }
  if( newCap > N ) {
    for( int i=N; i<newCap; ++i ) {
      tmp[ i ] = initial;
    }
  }

  arr = new int[ newCap ];

  for( int i=0; i<newCap; ++i ) {
    arr[ i ] = tmp[ i ];
  }
}

void print( int *arr, int N ) {
  for( int i=0; i<N; ++i ) {
    cout << arr[ i ];
    if( i != N-1 ) cout << " ";
  }
}

int main() {
  int arr[] = { 1, 2, 3, 4, 5 };

  print( arr, N );
  resize( arr, N, 5, 6 ); // line 37
  print( arr, N);
  resize( arr, N, 10, 1 ); // line 39
  print( arr, N );
  resize( arr, N, 3 ); // line 41
  print ( arr, N );

  return 0;
}

Could anyone help me? Thanks in advance.

@IonutHulub - that would defeat the purpose of the function, which is to change the size of the array that the pointer argument points to. — Pete Becker, Oct 02 '12 at 18:04
there is no such thing as a `*&arr`. what data type do you think that is? arr is a pointer so you can't call & on it, and even if you could, * undos the effect of & cause one is for reference and one if for dereference. — Ionut Hulub, Oct 02 '12 at 18:06
@IonutHulub actually there is. It is a reference to a pointer to an int and it allows you to make changes both to the pointer and to the value that it points to. — Rontogiannis Aristofanis, Oct 02 '12 at 18:08

Daniel Fischer · Accepted Answer · 2012-10-02T18:16:13.710

8

In main, you declare

int arr[] = { 1, 2, 3, 4, 5 };

an ordinary array, not an int*, which is what you would have to pass to resize.

And, although that is not technically necessary for your implementation of resize, since you never delete[] any allocated memory there - a space leak you should fix -, the pointer you pass to resize should point to a new-allocated memory block (that you should delete[] in resize).

edited Oct 02 '12 at 18:16

answered Oct 02 '12 at 18:03

Daniel Fischer

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it is impossible to resize a static array, so this isn't _quite_ the right advice, though you did identify the problem. – Mooing Duck Oct 02 '12 at 18:11
As is, passing any `int*` to `resize` will leak-but-work, `resize` will make it point to a different location (losing the possibly last reference to the old block). – Daniel Fischer Oct 02 '12 at 18:17

score 7 · Answer 2 · answered Oct 02 '12 at 18:03

7

arr is an array that's allocated on the stack. You can't change its size. If you need a resizable array, use a pointer and allocate the array with new.

answered Oct 02 '12 at 18:03

Pete Becker

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score 1 · Answer 3 · answered Oct 02 '12 at 18:09

An array is not a pointer. In particular, an array stays put. You cannot move it. In the context you are using the poibters you would also create memory leaks and out of bound accesses.

To create a resizable array, you will need to maintain a pointer which gets initially set up from a copy of the elements in you array. Personally, I wouldn't bother and use std::vector<int> (unless I'm working on my own standard C++ library and implement std::vector<T>).

score 0 · Answer 4 · edited May 23 '17 at 12:14

0

The reason you're getting the specific error you're getting is that arr will decay to a temporary int* as used. You then try and bind a non-const reference (the parameter) to the temporary, which is illegal. See here:

How come a non-const reference cannot bind to a temporary object?

edited May 23 '17 at 12:14

Community

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answered Oct 02 '12 at 18:15

Stuart Golodetz

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score 0 · Answer 5 · edited Oct 02 '12 at 20:37

0

When you say int& it means that an int variable that I can change its value, similarly when you say int*&, it means an int* variable that I can change its value. now look at your code can you change the value of arr. is it legal to say arr = new int[10]? you see it is not legal, in C++ term it is not an Lvalue(it can't sit in left side of equal operator) and all non-const references must be Lvalue

edited Oct 02 '12 at 20:37

Mooing Duck

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answered Oct 02 '12 at 18:38

BigBoss

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both `arr` and `new int[10]` can be on the right side of an equal operator, and all mutable references must be _lvalues_. – Mooing Duck Oct 02 '12 at 20:24
@MooingDuck thanks for your comment, it was a simple mistake!! I mean lvalue too. thanks again – BigBoss Oct 02 '12 at 20:29

reference to pointer to int causes error

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