Checking how many times a number has been entered - c++

Question

I'm trying to solve one of the questions on a task sheet I've received, to help me further in my understanding of C++ code from my class.
It kept showing 100000 in output after I entered the values. Where is that 1 coming from?
I know there are better ways to write code for this but I just want to know were is my problem.
The question is
(and I quote):
Write a program that:
Asks the user to enter 10 numbers between 1 and 5 into an array and displays the array on screen.
Creates a second array of size 5 and fills it with zeros.
Counts how many 1s, 2s, , … 5s have been entered into the first array and stores this number in the second array.
Displays the second array as shown in the example below.
Code:
int A1[10];
int A2[5] = { 0,0,0,0,0 };
int count = 10;
for (int i = 0; i < count; i++)
{
here:
cout << endl << i + 1 << "- enter a number between 1 and 5 for value : ";
cin >> A1[i];
if (A1[i] < 1 || A1[i]>5)
{
cout << "eror! enter a number between 1 and 5!";
goto here;
}
}
for (int i = 0; i < 10; i++)
{
for (int j = 1; j < 6; j++)
{
if (A1[i] = j)
{
A2[j - 1]++;
break;
}
}
}
for (int i = 0; i < 5; i++)
cout << A2[i];

Answer 1

The error is on the row 21 or 22, you are using a single = which is the assignment sign, inside the if statement, so you are overwriting the value of A[i] to the value of j, but want to check if the element of A[i] is equal to j... So you have to add a = in the if statement.

Answer 2

I don't recommend that you use goto:, it creates spaghetti code. you can put an i-- in your error clause, like so:
int temp;
for (int i = 0; i < count; i++){
cout << i + 1 << "- enter a number between 1 and 5 for value : " << endl;
cin >> temp;
if (temp >= 1 && temp <=5)
A1[i] = temp;
else
i--;
}
Also, if you want to compare 2 values, you should use the == operator. that is what's causing the problem in your second loop
like so:
for(int i = 0; i < count; i++){
for(int j = 1; j < 6; j++)
if(A1[i] == j)
A2[j-1]++;
}
This should work.