C++ concatenate two int arrays into one larger array

Question

Is there a way to take two int arrays in C++

int * arr1;
int * arr2;
//pretend that in the lines below, we fill these two arrays with different
//int values

and then combine them into one larger array that contains both arrays' values?

Answer 1

Use std::copy defined in the header <algorithm>. The args are a pointer to the first element of the input, a pointer to one past the last element of the input, and a pointer to the first element of the output. ( https://en.cppreference.com/w/cpp/algorithm/copy )

int * result = new int[size1 + size2];
std::copy(arr1, arr1 + size1, result);
std::copy(arr2, arr2 + size2, result + size1);

Just suggestion, vector will do better as a dynamic array rather than pointer

Answer 2

If you're using arrays, you need to allocate a new array large enough to store all of the values, then copy the values into the arrays. This would require knowing the array sizes, etc.

If you use std::vector instead of arrays (which has other benefits), this becomes simpler:

std::vector<int> results;
results.reserve(arr1.size() + arr2.size());
results.insert(results.end(), arr1.begin(), arr1.end());
results.insert(results.end(), arr2.begin(), arr2.end());

Answer 3

Another alternative is to use expression templates and pretend the two are concatenated (lazy evaluation). Some links (you can do additional googling):

http://www10.informatik.uni-erlangen.de/~pflaum/pflaum/ProSeminar/ http://www.altdevblogaday.com/2012/01/23/abusing-c-with-expression-templates/ http://aszt.inf.elte.hu/~gsd/halado_cpp/ch06s06.html

If you are looking for ease of use, try:

#include <iostream>
#include <string>

int main()
{
  int arr1[] = {1, 2, 3};
  int arr2[] = {3, 4, 6};

  std::basic_string<int> s1(arr1, 3);
  std::basic_string<int> s2(arr2, 3);

  std::basic_string<int> concat(s1 + s2);

  for (std::basic_string<int>::const_iterator i(concat.begin());
    i != concat.end();
    ++i)
  {
    std::cout << *i << " ";
  }

  std::cout << std::endl;

  return 0;
}

Answer 4

Here is the solution for the same-

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
void Concatenate(char *s1,char *s2)
{
    char s[200];
    int len1=strlen(s1);
    int len2=strlen(s2);
    int j;
    ///Define k to store the values on Kth address Kstart from 0 to len1+len2;
    int k=0;

        for(j=0;j<len1;j++)
        {
            s[k]=s1[j];
            k++;

        }
        for(j=0;j<len2;j++)
        {
            s[k]=s2[j];
            k++;
        }
        ///To place a null at the last of the concatenated string to prevent Printing Garbage value;
        s[k]='\n';

    cout<<s;
}

int main()
{
    char s1[100];
    char s2[100];
    cin.getline(s1,100);
    cin.getline(s2,100);
    Concatenate(s1,s2);

    return 0;
}

Hope it helps.

Answer 5

for (int i = 0; i< arraySize * 2; i++)
{
    if (i < aSize)
    {
        *(array3 + i) = *(array1 + i);
    }
    else if (i >= arraySize)
    {
        *(array3 + i) = *(array2 + (i - arraySize));
    }

}

This might help you along, it doesn't require vectors. I had a similar problem in my programming class. I hope this helps, it was required that I used pointer arithmetic. This assumes that array1 and array2 are initialized dynamically to the size "aSize"

Answer 6

Given int * arr1 and int * arr2, this program Concatenates in int * arr3 the elements of the both arrays. Unfortunately, in C++ you need to know the sizes of each arrays you want to copy. But this is no impediment to choose how many elements you want to copy from arr1 and how many from arr2.

#include <iostream>
using namespace std;

int main(){
int temp[] = {1,2,3,4};
int temp2[] = {33,55,22};
int * arr1, * arr2, *arr3;
int size1(4), size2(3); //size1 and size2 is how many elements you 
//want to copy from the first and second array. In our case all.
//arr1 = new int[size1]; // optional
//arr2 = new int[size2];

arr1=temp;
arr2=temp2;

arr3 = new int; 
//or if you know the size: arr3 = new int[size1+size2];

for(int i=0; i<size1+size2; i++){
    if (i<size1)
        arr3[i]=arr1[i];
    else
        arr3[i] = arr2[i-size1];
}
cout<<endl;
for (int i=0; i<size1+size2; i++) {
    cout<<arr3[i]<<", ";
}

}