Elementary C++, assigning arrays

[Lord Anoobis]

Homework Statement

Assuming that

array1

and

array2

are both arrays, why is it not possible to assign
the contents of

array2

to

array1

with the following statement?

array1 = array2;

Homework Equations

The Attempt at a Solution

This is a question I just saw in a book. As far as I can tell the given line doesn't work because the variables still need to be specified as arrays with [ ]. Anyway, I carried out a little experiment to see what happens when one tries something like this and I found...

#include <iostream>

int main ()
{
  int array1[5], array2[] = {1, 2, 3, 4, 5};
  array1[5] = array2[5];

  for (int i = 0; i < 5; i++)
  std::cout << array2[i] << "\t";
  std::cout << std::endl;

  for (int j = 0; j < 5; j++)
  std::cout << array1[j] << "\t";
  std::cout << std::endl;

  return 0;

}

Appears to assign random values to array1. Then I tried this...

#include <iostream>

int main ()
{
  int array2[] = {1, 2, 3, 4, 5};
  int array1[] = {array2[5]};

  for (int i = 0; i < 5; i++)
  std::cout << array2[i] << "\t";
  std::cout << std::endl;

  for (int j = 0; j < 5; j++)
  std::cout << array1[j] << "\t";
  std::cout << std::endl;

  return 0;

}

Which surprisingly yields a vastly improved (but still not quite correct) result. Even more bizarre is the result obtained when in the second program in line 6 we have

int array1[5] = {array2[5]}

. Why is this happening and how does one go about correctly setting one array equal to another? Without using a loop, that is.

 

[Lord Anoobis]

That didn't quite post the way I expected...

 

[phyzguy]

I don't think it is possible without looping through the index and assigning array1[j] = array2[j] for each j. What's wrong with doing this?

 

[DrClaude]

array1 = array2;

As far as I can tell the given line doesn't work because the variables still need to be specified as arrays with [ ].

You seem to be missing an important point: arrays variables are pointers.

  int array1[5], array2[] = {1, 2, 3, 4, 5};
  array1[5] = array2[5];

There is a big difference between declaring an array as array1[5], i.e., having a size of 5, as in the first line above, to referencing the sixth element array1[5], as in the second line. The second statement above will have unpredictable consequences. Do you see why?

  int array1[] = {array2[5]};

The presence of the curly brackets here doesn't do what you think it does.

Why is this happening and how does one go about correctly setting one array equal to another? Without using a loop, that is.

Look up memcpy.

 

[Lord Anoobis]

You seem to be missing an important point: arrays variables are pointers.

There is a big difference between declaring an array as array1[5], i.e., having a size of 5, as in the first line above, to referencing the sixth element array1[5], as in the second line. The second statement above will have unpredictable consequences. Do you see why?

I see it. There IS no sixth element. I'm not sure how that one slipped by.

 

[Lord Anoobis]

I don't think it is possible without looping through the index and assigning array1[j] = array2[j] for each j. What's wrong with doing this?

That's how I've been doing it, but I wanted to see what would happen if I tried it in the way I described.

 

[Mark44]

That's how I've been doing it, but I wanted to see what would happen if I tried it in the way I described.

What happens is a consequence of the difference in C between array1 (an address) and array1[k] (the element of the array at index k). The variable array1 is read-only, so it cannot appear on the left side of an assignment statement. In C parlance, "array1 is not a modifiable l-value."

 

[Lord Anoobis]

What happens is a consequence of the difference in C between array1 (an address) and array1[k] (the element of the array at index k). The variable array1 is read-only, so it cannot appear on the left side of an assignment statement. In C parlance, "array1 is not a modifiable l-value."

I'll be sure to remember that. Thanks.