Unexpected changing of C++ constant integer [closed] - c++

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The variable steady_counter is intialized as a constant integer.
cout << steady_counter;
So long as i have the above statement anywhere before the following code, the function runs as expected and checks if an integer input is or is not a runaround number.
The problem is that when the cout line is not present, the constant integer changes within the below if statements. I tested this by printing steady_counter before entering the if-else, and then after the if-else.
Without the cout line, steady_counter changes to a 4 digit number.
for (int i = 0; i < 10; i++)
{
if (CheckArr[i])
{
num_of_unique++;
}
}
if ((steady_counter == num_of_unique) & (final == NumArr[0]) )
{
return true;
}
else
{
return false;
}
}
Any idea what's going on? Why do I require a cout line to maintain the constant integer steady_counter?

Answer 1

One obvious problem:
for (int i = counter; i > 0; i --)
NumArr[i] = -1;
This covers values from 1 to counter inclusive; while valid indexes for NumArr are from 0 to counter-1 inclusive. So you write outside the array, corrupting something else; possibly another local variable.
Either correct the off-by-one error in the index
NumArr[i-1] = -1;
or use a more canonical loop
for (int i = 0; i < counter; ++i)
or, for more of a C++ flavour,
std::fill(NumArr, NumArr+counter, -1);
There are likely to be further errors, which are better found by using your debugger than by asking people to read through all your code.