C++ Counting Matched Numbers help

[scooter28]

I'm trying to get an output as this:

Enter 6 numbers into the first array: 5 3 8 2 9 4
Enter 6 numbers into the second array: 1 4 5 2 7 6
There are 3 matched numbers between the two arrays.

Note that I want it to display the amount of matched numbers, not the value of the number that is matched that it seems to be doing here. Would appreciate the help.

#include <iostream>
#include <cstdlib>

using namespace std;

int main() 
{
    int p[6], p1[6], num = 0, p2[] = {};
    cout << "Enter 6 numbers into the first array: ";
    
    for (int a=0;a<6;a++)
    {
    	cin >> p[a];
    }
    
    cout << "Enter 6 numbers into the second array: ";
    
    for (int a=0;a<6;a++)
    {
    	cin >> p1[a];
    }
    
    for (int i = 0 ; i < 6; ++i) 
    {
        for (int j = 0 ; j < 6; ++j) 
        {
            
    	if (p[i]==p1[j]) 
    	{
    		p2[num++] = p[j];
        }
    }
    
    	cout << "There are " << num << " matched numbers between the two arrays.";
    	return 0;
    }
}

[Zaap]

Missing one '}' :

for (int i = 0 ; i < 6; ++i) 
    {
        for (int j = 0 ; j < 6; ++j) 
        {
            
    	    if (p[i]==p1[j]) 
    	    {
    		p2[num++] = p[j];
            }
        } //this one was missing
}

Remove the last one.

[scooter28]

@Zaap

The program displays properly however the problem is that I'm not getting the output that I want, when I try to get the matched values it is wrong.

[Zaap]

Because you end the program before the end of the loop, and this beacause a '}' is missplaced.

[scooter28]

#include <iostream>
#include <cstdlib>

using namespace std;

int main() 
{
    int p[6], p1[6], num = 0, p2[] = {};
    cout << "Enter 6 numbers into the first array: ";
    
    for (int a=0;a<6;a++)
    {
    	cin >> p[a];
    }
    
    cout << "Enter 6 numbers into the second array: ";
    
    for (int a=0;a<6;a++)
    {
    	cin >> p1[a];
    }
    
    for (int i = 0 ; i < 6; ++i) 
    {
        for (int j = 0 ; j < 6; ++j) 
        {
            
    	    if (p[i]==p1[j]) 
    	    {
    	        
    		p2[num++] = p[j];
    		
            }
            
        } //this one was missing
    }   
    
    cout << "There are " << num << " matched numbers between the two arrays.";
    return 0;
}

When I input "5 3 8 2 9 4" into the first array and "1 4 5 2 7 6" into the second it displays the correct matched value of 3.

However if I input for example "9 8 7 6 5 4" into both the first and second array it displays as 12 matched values when it should be 6. Thoughts?

[lastchance]

Do away with the p2[] array completely. (What you are doing with it may be legal in some languages, but not C++; turn the compiler error messages up and it should tell you that.) You are just incrementing num, so simply replace line 32 by num++;

However, you still need to think what you will do about mismatching numbers of repeats in your arrays. What if the first array is 1 1 1 1 1 1 and the second is 1 2 3 4 5 6 (or vice versa)?

[scooter28]

Not sure what you mean when you say do away with the p2[] array. How do I fix this to make the code work? Sorry, I am a beginner. For mismatching numbers in that example I would like them to state as 1 matched number because both arrays would have '1'.

[lastchance]

Not sure what you mean when you say do away with the p2[] array.

Just remove it! Entirely!

Change line 32 to num++;

When you have got that to work then you can start worrying about repeats.

[scooter28]

Right, so the repeats, I do not know how to deal with. >_>

[lastchance]

For the repeats, one possibility is just to store a parallel 6-element array of bools indicating which array elements (of p1[]) had already been matched up and hence couldn't be compared with again.

I presume your code is now working without repeats?

[scooter28]

Yes, it is working without repeats.

[lastchance]

Well, to prevent an element of the inner-loop array (p2[]) being matched twice you can declare an array of bools:
bool used[6] = { false }; // initialise a 6-element array of bools to false
and then your matching later would be

    	    if (!used[j] && p[i]==p1[j]) 
    	    {
          		num++;
         		used[j] = true;
            }

There are other ways to do the problem ... and 6 is a "magic number" ... and there's a slightly simpler form of initialisation and ... tomorrow's another day!

[scooter28]

Thanks for the help.