Comparing arrays for equality in C++

Question

Can someone please explain to me why the output from the following code is saying that arrays are not equal?

int main()
{

    int iar1[] = {1,2,3,4,5};
    int iar2[] = {1,2,3,4,5};

    if (iar1 == iar2)
        cout << "Arrays are equal.";
    else
        cout << "Arrays are not equal.";

    return 0;   
}

Use `std::array` or `std::vector`. C arrays have no single advantage and they only bring pain and sorrow. No excuses. — daknøk, Oct 12 '12 at 20:14
No it doesn't but it's there for C interop. And no, converting C arrays to std::arrays, just to check for equality, is completely unacceptable. It involves copying. When `std::array_view` is in the standard we'll have a sensible way of comparing raw arrays resulting from C interop. — emlai, Apr 20 '16 at 00:47
Your declared variables are pointers to the first int elements in two seperate arrays in two distinct locations in memory. They are not equal (the addresses are different) and so the conditional test evaluates to false. — Reno, Feb 20 '20 at 10:11

Praetorian · Accepted Answer · 2016-12-29T06:51:14.073

88

if (iar1 == iar2)

Here iar1 and iar2 are decaying to pointers to the first elements of the respective arrays. Since they are two distinct arrays, the pointer values are, of course, different and your comparison tests not equal.

To do an element-wise comparison, you must either write a loop; or use std::array instead

std::array<int, 5> iar1 {1,2,3,4,5};
std::array<int, 5> iar2 {1,2,3,4,5};

if( iar1 == iar2 ) {
  // arrays contents are the same

} else {
  // not the same

}

edited Dec 29 '16 at 06:51

answered Oct 12 '12 at 20:18

Praetorian

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+1 for the only answer with the word "decay" (or anything else explicitly saying so). – chris Oct 12 '12 at 20:19
5

"... you must either write a loop...." Better would be to use std::equal. Not to detract from your point that using raw arrays is the main problem. – Ben Oct 19 '12 at 17:40
For people like me who got something like "implicit instantiation of undefined template ..." when trying to use std::array, try #import or #include . – Golden Thumb Mar 01 '18 at 03:36
2

Good explanation of "decay": https://stackoverflow.com/a/1461449/108238 – schoetbi Sep 30 '19 at 05:51

score 66 · Answer 2 · answered Oct 13 '12 at 07:25

66

Since nobody mentioned it yet, you can compare arrays with the std::equal algorithm:

int iar1[] = {1,2,3,4,5};
int iar2[] = {1,2,3,4,5};

if (std::equal(std::begin(iar1), std::end(iar1), std::begin(iar2)))
    cout << "Arrays are equal.";
else
    cout << "Arrays are not equal.";

You need to include <algorithm> and <iterator>. If you don't use C++11 yet, you can write:

if (std::equal(iar1, iar1 + sizeof iar1 / sizeof *iar1, iar2))

answered Oct 13 '12 at 07:25

InstallingXubuntuRightNow

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An interesting fact is that the above conditionals still hold true even if iar2 is {1,2,3,4,5,6}. – Golden Thumb Mar 01 '18 at 03:57
2

The beauty of this is that the exact same syntax also works for `std::vector` and `std::array` if one day you regain your sanity and want to convert the code to use them instead of C arrays. – Ciro Santilli新疆棉花TRUMP BAN BAD Oct 26 '18 at 10:36
1

How does this manage to avoid overrunning the end of `iar2` if `iar1` is bigger than `iar2`? Shouldn't you be passing `std:end(iar2)` in there too? – David Given Mar 04 '19 at 19:31

score 15 · Answer 3 · answered Oct 12 '12 at 20:13

15

You're not comparing the contents of the arrays, you're comparing the addresses of the arrays. Since they're two separate arrays, they have different addresses.

Avoid this problem by using higher-level containers, such as std::vector, std::deque, or std::array.

answered Oct 12 '12 at 20:13

Fred Larson

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It's important to note that these containers have implemented their own `==` operator that performs this check. – tadman Oct 12 '12 at 20:23

score 6 · Answer 4 · answered Oct 12 '12 at 20:13

6

Array is not a primitive type, and the arrays belong to different addresses in the C++ memory.

answered Oct 12 '12 at 20:13

Paul S.

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score 5 · Answer 5 · answered Jul 11 '19 at 13:06

Nobody mentions memcmp? This is also a good choice.

/* memcmp example */
#include <stdio.h>
#include <string.h>

int main ()
{
  char buffer1[] = "DWgaOtP12df0";
  char buffer2[] = "DWGAOTP12DF0";

  int n;

  n=memcmp ( buffer1, buffer2, sizeof(buffer1) );

  if (n>0) printf ("'%s' is greater than '%s'.\n",buffer1,buffer2);
  else if (n<0) printf ("'%s' is less than '%s'.\n",buffer1,buffer2);
  else printf ("'%s' is the same as '%s'.\n",buffer1,buffer2);

  return 0;
}

Ref: http://www.cplusplus.com/reference/cstring/memcmp/

The problem with the logic of this code is the `sizeof(buffer1) ` in memcpm — luca, Mar 21 '21 at 10:45

Saurav Sahu · Answer 6 · 2017-09-29T05:11:18.870

If you are reluctant to change your existing code to std::array, then use a couple of methods instead which takes non-type template arguments :

//Passed arrays store different data types
template <typename T, typename U, int size1, int size2>
bool equal(T (&arr1)[size1], U (&arr2)[size2] ){
    return false;
}

//Passed arrays store SAME data types
template <typename T, int size1, int size2>
bool equal(T (&arr1)[size1], T (&arr2)[size2] ){
    if(size1 == size2) {
        for(int i = 0 ; i < size1; ++i){
            if(arr1[i] != arr2[i]) return false;
        }
        return true;
    }
    return false;
}

Here is the demo. Note that, while calling, we just need to pass the array variables e.g. equal(iar1, iar2) in your case, no need to pass the size of arrays.

score 1 · Answer 7 · answered Oct 12 '12 at 20:15

1

You are comparing the addresses instead of the values.

answered Oct 12 '12 at 20:15

Rahul Tripathi

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score 1 · Answer 8 · answered May 02 '20 at 00:03

1

Both store memory addresses to the first elements of two different arrays. These addresses can't be equal hence the output.

answered May 02 '20 at 00:03

Vachaspati Mishra

11
1

almost correct @Vachaspati -- they could be equal (for example, after the command `iar1 = iar2;`) but in this case, straight after initialisation, they won't be. – alle_meije May 02 '20 at 05:27

slugonamission · Answer 9 · 2012-10-12T20:28:28.050

0

Right. In most, if not all implementations of C, the array identifier can be implicitly casted to a pointer to the first element (i.e. the first element's address). What you're doing here is comparing those addresses, which is obviously wrong.

Instead, you need to iterate over both arrays, checking each element against each other. If you get to the end of both without a failure, they're equal.

edited Oct 12 '12 at 20:28

answered Oct 12 '12 at 20:14

slugonamission

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In *none* is the array identifier actually the address of the first element. The array identifier is actually the array. In `int arr[6]`, `arr` refers to a value of type `int[6]`. That value is implicitly convertible to `int*`, with the value `&arr[0]` (often called decaying). But an array is not "actually" a pointer. – GManNickG Oct 12 '12 at 20:25

vladinkoc · Answer 10 · 2012-10-17T11:01:03.403

0

When we use an array, we are really using a pointer to the first element in the array. Hence, this condition if( iar1 == iar2 ) actually compares two addresses. Those pointers do not address the same object.

edited Oct 17 '12 at 11:01

answered Oct 12 '12 at 21:49

vladinkoc

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score 0 · Answer 11 · answered Nov 30 '20 at 23:27

0

If you are willing to use std::array instead of built-in arrays, you may use:

std::array<int, 5> iar1 = {1,2,3,4,5};
std::array<int, 5> iar2 = {1,2,3,4,5};

if (iar1 == iar2)

answered Nov 30 '20 at 23:27

R Sahu

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Comparing arrays for equality in C++

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