Find similar numbers - array inconsistency - c++

Question

I'm trying to solve a simple beginner exercise, I compare two arrays and find the numbers that appear in both. The result is put into another array called result. For whatever reason the result should contain "2 44 55" but it shows "2 1 10 10". What did I do wrong?
#include <iostream>
void common_elements(int array_1[], int array_2[]){
int result[] {0};
int counter{0};
for (int i{0}; i < 10; i++){
for (int j{0}; j < 10; j++){
if (array_2[i] == array_1[j]){
result[counter] = array_1[j];
counter++;
}
}
}
if (counter == 0) {
std::cout << "There are 0 common elements";
} else {
std::cout << "There are " << counter << " common elements they are : ";
for (int k{0}; k < counter; k++){
std::cout << result[k] << " ";
}
}
}
int main(){
int data1[] {1,2,4,5,9,3,6,7,44,55};
int data2[] {11,2,44,45,49,43,46,47,55,88};
common_elements(data1,data2);
return 0;
}
I'm confused because when I std::cout the numbers during examination (two nested loops), the result is correct.

Answer 1

It is just containing one single element, because the compiler deduces the length to 1. This would mean the 2 is valid, all other values are "out of bound". Thanks for your help. I repaired it to
int result[10]{0};
Now it's working, thanks a lot.
Btw: It is a C++ course but it starts from the very beginning. That why this looks like C.

Answer 2

For starters the function should be declared at least like
std::vector<int> common_elements( const int array_1[], size_t n1,
const int array_2[], size_t n2 );
Using the magic number 10 within the function as
for (int i{0}; i < 10; i++){
does not make sense. The function can be called for arrays that have different numbers of elements.
This declaration of an array
int result[] {0};
declares an array with only one element that also does not make sense.
Also the function should not output any message. It is the caller of the function that decides whether to output a message. The function should return a sequence of common elements of two arrays.
The function can be defined the following way
#include <vector>
#include <iterator>
#include <functional>
//...
std::vector<int> common_elements( const int array_1[], size_t n1,
const int array_2[], size_t n2 )
{
std::vector<int> v;
if ( n1 != 0 && n2 != 0 )
{
if ( n2 < n1 )
{
std::swap( n1, n2 );
std::swap( array_1, array_2 );
}
for ( size_t i = 0; i < n1; i++ )
{
size_t count = 1;
for ( size_t j = 0; j < i; j++ )
{
if ( array_1[j] == array_1[i] ) ++count;
}
for ( size_t j = 0; count != 0 && j < n2; j++ )
{
if ( array_2[j] == array_1[i] )
{
--count;
}
}
if ( count == 0 ) v.push_back( array_1[i] );
}
}
return v;
}
And in main the function is called like
int data1[] {1,2,4,5,9,3,6,7,44,55};
int data2[] {11,2,44,45,49,43,46,47,55,88};
auto v = common_elements( data1, std::size( data1 ), data2, std::size( data2 ) );
if ( std::size( v ) == 0 )
{
std::cout << "There are 0 common elements";
}
else
{
std::cout << "There are " << std::size( v ) << " common elements they are : ";
for ( const auto &item : v )
{
std::cout << item << ' ';
}
std::cout << '\n';
}
Instead of the vector you could use std::map<intg, size_t>. In this case the container will contain how many times a common number is encountered in the both arrays.
Here is a demonstration program.
#include <iostream>
#include <vector>
#include <iterator>
#include <functional>
std::vector<int> common_elements( const int array_1[], size_t n1,
const int array_2[], size_t n2 )
{
std::vector<int> v;
if (n1 != 0 && n2 != 0)
{
if (n2 < n1)
{
std::swap( n1, n2 );
std::swap( array_1, array_2 );
}
for (size_t i = 0; i < n1; i++)
{
size_t count = 1;
for (size_t j = 0; j < i; j++)
{
if (array_1[j] == array_1[i]) ++count;
}
for (size_t j = 0; count != 0 && j < n2; j++)
{
if (array_2[j] == array_1[i])
{
--count;
}
}
if (count == 0) v.push_back( array_1[i] );
}
}
return v;
}
int main()
{
int data1[]{ 1,2,4,5,9,3,6,7,44,55 };
int data2[]{ 11,2,44,45,49,43,46,47,55,88 };
auto v = common_elements( data1, std::size( data1 ), data2, std::size( data2 ) );
if (std::size( v ) == 0)
{
std::cout << "There are 0 common elements";
}
else
{
std::cout << "There are " << std::size( v ) << " common elements they are : ";
for (const auto &item : v)
{
std::cout << item << ' ';
}
std::cout << '\n';
}
}
The program output is
There are 3 common elements they are : 2 44 55