Frequency array of c++

Question

#include <iostream>

using namespace std;
int main()
{
    int arr,i=0,frequancy,y,X,l;
    
    int V[arr];
    cout<<"Enter the size of an array: ";
    cin>>arr;
    
    cout<<"Enter all "<<arr<<" elements in array: ";
    for(i=0;i<=arr-1;i++)
    cin>>V[i];
    
    int a[arr]={0};
    
    cout<<"Frequency of all "<<arr<<" elements in array:"<<endl;
        
    for(i=0;i<arr;i++)
    {
        int count=0;
        if (a[i]!=1)
        {
            for (int j=0;j<arr;j++)
            {
                if(V[i]==V[j])
                {
                    count++;
                    a[j]=1;
                }
            }
            if (count>1||count==1)
            {
                cout<<V[i]<<" occurs "<<count<<" times "<<endl;
            }
        }
    }
return 0;
}

//this is the answer
Enter the size of an array: 9
Enter all 9 elements in array: 1
1
2
3
1
5
3
12
1
Frequency of all 9 elements in array:
1 occurs 4 times
2 occurs 1 times
3 occurs 2 times
5 occurs 1 times
12 occurs 1 times

So there are a few questions that I don't understand that well and I need someone to explain what is going on with it, please.

1. I don't understand why we have to initialize this(int a[arr]={0}) to 0, I asked my prof about it be he said it has to be like that. which doesn't answer my question:'C.

2. I have a problem with the bottom part of the code, I kinda understand a bit but if someone can help me that would help a lot Thank you!!!!

Answer 1

int arr/*...*/;
int V[arr];
cin>>V[i];

This program is badly broken.

Problem 1: The size of an array variable must be a compile time constant in C++. arr is not a compile time constant value. Violating this rule makes the program ill-formed, which means that compilers aren't required to compile it. You should avoid this.

Solution: If you need an array whose size is determined at runtime, then you should allocate the array dynamically. Simplest way to achieve that is to use the std::vector template from the standard library.

---

Problem 2: The behaviour of a program that reads an indeterminate value is undefined. Undefined behaviour means that there are no guarantees about the behaviour of the program whatsoever. This is very bad and should be avoided.

arr has an indeterminate value when you use it. The behaviour of the program is undefined. I repeat: This is very bad.

C++ is not a "data-flow" language. You cannot simply use a variable and initialise it after the usage. The program won't automatically stop, create a thread and wait for the later initialisation to proceed further.

Solution: Initialise before use.

---

I don't understand why we have to initialize this(int a[arr]={0}) to 0

Later on in the program, the values of that array are read (a[i]!=1). If you don't initialise the values before using them, then the behaviour of the program is undefined. See the previous paragraph.

Presumably, the elements of the array are initialised to zero in order to avoid that undefined behaviour.

Note that it is redundant to initialise the first element of the array explicitly to zero, and value initialise the rest, because those achieve the same thing. Thus, a simpler way to achieve the same is to value initialise all of the elements by not providing an initialiser for any of them:

constexpr int arr = 42;
int a[arr] {};

---

I have a problem with the bottom part of the code

The first step to solve that problem is to figure out why you think that you have a problem, and what kind of problem it could be.