144

I know that Java has a function System.arraycopy(); to copy an array. I was wondering if there is a function in C or C++ to copy an array. I was only able to find implementations to copy an array by using for loops, pointers, etc. But is there a library function that I can use to copy an array?

9
  • 8
    man memmove and man memcpy Commented Apr 22, 2013 at 1:06
  • 36
    Don't use memcpy, use std::copy. If your type has a meaningful copy constructor then memcpy will do the wrong thing. Commented Apr 22, 2013 at 1:17
  • 13
    Are you actually trying to learn C and C++ at the same time? They are very different languages. Commented Apr 22, 2013 at 1:21
  • 9
    I'm saying this because no one has mentioned it before: In C++ you should use std::vector in almost all cases. There are cases where other containers are usefull too, but i most cases std::vector will be the best option. Don't use raw arrays in C++ and try to avoid std::array unless neccessary. Commented Apr 22, 2013 at 14:27
  • 4
    Pick C or C++. They're very distinct. Commented Dec 6, 2016 at 23:18

14 Answers 14

201

Since you asked for a C++ solution...

#include <algorithm>
#include <iterator>

const int arr_size = 10;
some_type src[arr_size];
// ...
some_type dest[arr_size];
std::copy(std::begin(src), std::end(src), std::begin(dest));
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8 Comments

erreur: ‘begin’ is not a member of ‘std’
@David天宇Wong: You're using an old compiler or you didn't include the header. It's in <iterator>
Maybe you should say which includes you need. Is copy in <algorithm> ?
I only ask because if you had to explain that std::begin was in <iterator> that means people are not googling std::begin to find out what header it is in so they are unlikely to google std::copy for the same reason. It makes a better answer if nobody has to ask stuff that is only answered in comments. Maybe I should have just edited the answer instead of prompting you to do it.
you can change begin(src) into src and end(src) into src+arr_size
|
117

Since C++11, you can copy arrays directly with std::array:

#include <array>
std::array<int, 4> A = {10, 20, 30, 40};
std::array<int, 4> B = A; // Copy array A into array B

4 Comments

@XAleXOwnZX: The same you would any other type that supports copy assignment. B = A.
I think the function is copy A's memory address to B. If I try to copy A's item values to B without changing B's memory address. How can I do that?
@Aaron_Lee: I've just tested it, and it genuinely copies the elements into an array in a separate memory location. The array class evidently has its own overload for the assignment operator.
You guys realize that no assignment operators appeared in those two lines of code, right? That's the copy constructor that is being called even though the notation looks like the assignment operator.
108

As others have mentioned, in C you would use memcpy. Note however that this does a raw memory copy, so if your data structures have pointer to themselves or to each other, the pointers in the copy will still point to the original objects.

In C++ you can also use memcpy if your array members are POD (that is, essentially types which you could also have used unchanged in C), but in general, memcpy will not be allowed. As others mentioned, the function to use is std::copy.

Having said that, in C++ you rarely should use raw arrays. Instead you should either use one of the standard containers (std::vector is the closest to a built-in array, and also I think the closest to Java arrays — closer than plain C++ arrays, indeed —, but std::deque or std::list may be more appropriate in some cases) or, if you use C++11, std::array which is very close to built-in arrays, but with value semantics like other C++ types. All the types I mentioned here can be copied by assignment or copy construction. Moreover, you can "cross-copy" from opne to another (and even from a built-in array) using iterator syntax.

This gives an overview of the possibilities (I assume all relevant headers have been included):

#include <vector>
int main()
{
  // This works in C and C++
  int a[] = { 1, 2, 3, 4 };
  int b[4];
  memcpy(b, a, 4*sizeof(int)); // int is a POD

  // This is the preferred method to copy raw arrays in C++ and works with all types that can be copied:
  std::copy(a, a+4, b);

  // In C++11, you can also use this:
  std::copy(std::begin(a), std::end(a), std::begin(b));

  // use of vectors
  std::vector<int> va(a, a+4); // copies the content of a into the vector
  std::vector<int> vb = va;    // vb is a copy of va

  // this initialization is only valid in C++11:
  std::vector<int> vc { 5, 6, 7, 8 }; // note: no equal sign!

  // assign vc to vb (valid in all standardized versions of C++)
  vb = vc;

  //alternative assignment, works also if both container types are different
  vb.assign(vc.begin(), vc.end());

  std::vector<int> vd; // an *empty* vector

  // you also can use std::copy with vectors
  // Since vd is empty, we need a `back_inserter`, to create new elements:
  std::copy(va.begin(), va.end(), std::back_inserter(vd));

  // copy from array a to vector vd:
  // now vd already contains four elements, so this new copy doesn't need to
  // create elements, we just overwrite the existing ones.
  std::copy(a, a+4, vd.begin());

  // C++11 only: Define a `std::array`:
  std::array<int, 4> sa = { 9, 10, 11, 12 };

  // create a copy:
  std::array<int, 4> sb = sa;

  // assign the array:
  sb = sa;
}

5 Comments

why are you writing std all the time instead of just using the namespace of std ???
can't you do memcpy(b, a, sizeof a); too?
A Java array is more like std:array than like std::vector, because it has a fixed size.
Does std::begin() buy you anything in your second example of std::copy?
@wcochran: Yes: Should you ever decide to rewrite the code to use an actual container instead of a built-in array, you will not have to change that line. The less code you need to change, the less bugs you will introduce (and the less work you will have).
23

Use memcpy in C, std::copy in C++.

2 Comments

For a small array I assume neither incur any function call overhead (memcpy should be a compiler intrinsic)?
@Mehrdad I didn't say otherwise; just making a comment. And I didn't downvote it, either.
19

You can use the memcpy(),

void * memcpy ( void * destination, const void * source, size_t num );

memcpy() copies the values of num bytes from the location pointed by source directly to the memory block pointed by destination.

If the destination and source overlap, then you can use memmove().

void * memmove ( void * destination, const void * source, size_t num );

memmove() copies the values of num bytes from the location pointed by source to the memory block pointed by destination. Copying takes place as if an intermediate buffer were used, allowing the destination and source to overlap.

7 Comments

Gonna have to -1 here. The OP is specifically asking for a C++ solution and this answer says nothing of the situations in which memcpy is wrong, i.e., copying bytes is insufficient.
@EdS. In case you haven't noticed, the OP is now asking for a C or C++ solution.
When I answered the OP was asking for C/C++ solutions, though I personally believe "C/C++" is insult to both the languages. :)
fair enough, I didn't realize it was changing. Maybe it even said that earlier and I missed it. -1 removed.
@EdS. It always said "a function in C or C++".
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18

I like the answer of Ed S., but this only works for fixed size arrays and not when the arrays are defined as pointers.

So, the C++ solution where the arrays are defined as pointers:

#include <algorithm>
...
const int bufferSize = 10;
char* origArray, newArray;
std::copy(origArray, origArray + bufferSize, newArray);

Note: No need to deduct buffersize with 1:

  1. Copies all elements in the range [first, last) starting from first and proceeding to last - 1

See: https://en.cppreference.com/w/cpp/algorithm/copy

2 Comments

But how do we ensure that no "overlapping" will occur ?
I don't think you can, just don't use raw pointers ;-)
12

In C you can use memcpy. In C++ use std::copy from the <algorithm> header.

Comments

6

I give here 2 ways of coping array, for C and C++ language. memcpy and copy both ar usable on C++ but copy is not usable for C, you have to use memcpy if you are trying to copy array in C.

#include <stdio.h>
#include <iostream>
#include <algorithm> // for using copy (library function)
#include <string.h> // for using memcpy (library function)


int main(){

    int arr[] = {1, 1, 2, 2, 3, 3};
    int brr[100];

    int len = sizeof(arr)/sizeof(*arr); // finding size of arr (array)

    std:: copy(arr, arr+len, brr); // which will work on C++ only (you have to use #include <algorithm>
    memcpy(brr, arr, len*(sizeof(int))); // which will work on both C and C++

    for(int i=0; i<len; i++){ // Printing brr (array).
        std:: cout << brr[i] << " ";
    }

    return 0;
}

1 Comment

using namespace std; is a bad practice. Don't ever use it. Instead of cplusplus.com, please use cppreference.com, which includes much better and up-to-date documentation of the standard. The stdio.h equivalent in C++ is cstdio, use that instead. Additionally, the code is fairly non-idiomatic C++. I think it'd be much clearer if you had showcased separate solutions for C and C++.
6

Just include the standard library in your code.

#include<algorithm>

Array size will be denoted as n

Your old Array

int oldArray[n]={10,20,30,40,50};

Declare New Array in which you have to copy your old array value

int newArray[n];

Use this

copy_n(oldArray,n,newArray);

Comments

3

in C++11 you may use Copy() that works for std containers

template <typename Container1, typename Container2>
auto Copy(Container1& c1, Container2& c2)
    -> decltype(c2.begin())
{
    auto it1 = std::begin(c1);
    auto it2 = std::begin(c2);

    while (it1 != std::end(c1)) {
        *it2++ = *it1++;
    }
    return it2;
}

1 Comment

It would be a better idea to pull the std::end out of the loop, for performance reasons (c1 doesn't change or invalidate iterators during the loop,so it's unnecessary to re-calculate the end).
3

Firstly, because you are switching to C++, vector is recommended to be used instead of traditional array. Besides, to copy an array or vector, std::copy is the best choice for you.

Visit this page to get how to use copy function: http://en.cppreference.com/w/cpp/algorithm/copy

Example:

std::vector<int> source_vector;
source_vector.push_back(1);
source_vector.push_back(2);
source_vector.push_back(3);
std::vector<int> dest_vector(source_vector.size());
std::copy(source_vector.begin(), source_vector.end(), dest_vector.begin());

Comments

2

C++ 20

You can use std::ranges::copy

as simple, as std::ranges::copy(a, b);

#include <algorithm> // ranges::copy, ranges::copy_if
#include <iostream>  // cout
#include <iterator>  // ostream_iterator, begin, end
#include <numeric>   // iota

int main()
{
    float source[10];
    std::iota(std::begin(source), std::end(source), 0);

    float destination[10];

    std::ranges::copy(source, destination);

    std::cout << "destination contains: ";
    std::ranges::copy(destination, std::ostream_iterator<float>(std::cout, " "));
    std::cout << '\n';

    std::cout << "odd numbers in destination are: ";

    std::ranges::copy_if(destination, std::ostream_iterator<float>(std::cout, " "),
                         [](int x) { return (x % 2) == 1; });
    std::cout << '\n';
}

https://coliru.stacked-crooked.com/view?id=1a20680b445b5749

5 Comments

Found a really interesting thing. if you make the arrays with int arr[num] syntax std::ranges::copy works fine but if you heap allocate with int* arr = new int[num] it throws a really weird error: no instance of overloaded function "std::ranges::_Copy_fn::operator()" matches the argument list
@ThomasBourne in the first case arr is the reference to fixed sized array, so it can infer array size from type. In the second case arr is just a pointer to single int (first element).
@ThomasBourne you can use 2 arg iterator overload (cause pointers are some kind of iterstor), i.e. std::ranges::copy(source, source + num, destination);
ahhh thank you for this, I didnt realise fundimentally they were different in the way they were stored. That is really interesting and useful thank you mate!
@ThomasBourne the difference not in the storage, but in the type of variable. In the first case arr is of an array type (int [10]), in the second arr is just int* (cause operator new only returns pointers). In C++ array may implicitly decay to pointer, but they are not same things.
1

Try this :

  1. Create an empty array.
  2. Insert the elements.
  3. Create a duplicate empty array of the same size.
  4. Start for i=0 to i=array length.
    5. newarray[i]=oldarray[i] (only for C++)

C++ program

#include<iostream>
using namespace std;
int main()
{
        int initA[100],finA[100],i,size;
        cout<<"Input the size of the array : ";
        cin>>size;
        cout<<"Input the elements of the first array";
        for(i=0;i<size;i++)
        {
               cin>>initA[i];
        }
        for(i=0;i<size;i++)
        {
             finA[i]=initA[i];
        }
        cout<<"The final array is\n";
        for(i=0;i<size;i++)
               cout<<finA[i]<<" ";
        return 0;
}

1 Comment

I don't see why you would write a loop manually instead of just using the standard copy or memcpy methods.
1

It's interesting that while you can easily copy structs in C/C++, the same straightforward approach doesn't work for arrays. An amusing workaround involves casting an array to a struct pointer, making it possible to copy arrays, as demonstrated below:

    int a[10] = {1, 2, 3};
    int b[10];

    {
        typedef struct {char _[sizeof(a)];} _;
        *(_*)b = *(_*)a;
    }

However, this solution may seem unattractive due to its unconventional appearance. To address this, we can encapsulate the operation within a macro:

#include <assert.h>
#include <stdio.h>

#define ARRAY_ASSIGN(x, y) \
{ assert(sizeof(x) == sizeof(y)); \
  typedef struct {char _[sizeof(x)];} _; *(_*)(x) = *(_*)(y);\
}


int
main(void)
{
    int a[10] = {1, 2, 3};
    int b[10];

    ARRAY_ASSIGN(b, a); // b = a

    for (size_t i = 0; i < sizeof(b)/sizeof(b[0]); ++i) {
        printf("%d ", b[i]);
    }
    printf("\n");
}

🤣🤣🤣

Comments

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