" 'X' not declared in this scope " error - c++

Question

I'm very new to programming, I'm just in the 2nd sem of my first year in college so please go easy on the technical terms. We were asked to make a program that reads 10 integers from a file to comprise a list and asks the user to input an integer 'N'. If 'N' is in the list the program should display "FOUND" and "NOT FOUND" if otherwise. I'm getting an error in main about the arguments, it says that 'V', 'N', and 'F' in the function calls "was not declared in this scope".
#include<iostream>
#include<fstream>
using namespace std;
int fRead();
int iRead();
bool search(int, int);
void display(bool);
int main() {
fRead();
iRead();
search(V, N);
display(F);
return 0;
}
int fRead() {
int V[10], c;
ifstream fin;
fin.open("lab02.in");
for(c=0; c<10; c++)
fin >> V[10];
fin.close();
return V[10];
}
int iRead() {
int N;
cout << "Input an integer: ";
cin >> N;
return N;
}
bool search(int V[10], int N) {
bool F = false;
if(V[10] == N)
F = true;
return F;
}
void display(bool F) {
if(F == true)
cout << "\nFOUND" << endl;
else
cout << "\nNOT FOUND" << endl;
}

Answer 1

Local variable (those declared inside a function) are only visible to the block (those things delimited by { and }) where they are declared. If you want to use different functions for various operations you'll need to pass the variables as arguments to the corresponding functions.
As an aside, you should always verify that your read operation was successful before using the result, e.g.:
int N(-1);
if (!(std::cin >> N)) {
std::cout << "ERROR: failed to read integer\n";
}

Answer 2

the integers V N and F are basically described in other functions.
To solve this problem you should declare them in main
the program should go as follows
void f_read(v[]);
void i_read(int &);
bool bool(int,int);
void disp(bool);
void main()
{
int v[10],n;
bool f;
f_read(v);
i_read(n);
f=bool(v,n);
disp(f);
}
void fRead()
{
int c;
ifstream fin;
fin.open("lab02.in");
for(c=0; c<10; c++)
fin >> V[c];
fin.close();
}
void iRead(int &n)
{
cout << "Input an integer: ";
cin >> N;
}
bool search(int V[10], int N)
{
bool F = false;
int i;
for(i=0;i<=9;i++)
if(V[i] == N)
{
F = true;
return F;
}
}
void display(bool F)
{
if(F == true)
cout << "\nFOUND" << endl;
else
cout << "\nNOT FOUND" << endl;
}
so basically you need to pass n as a reference variable and arrays by default are passed as reference.
and also you were looking for v[10]=f ,which would have given you another array, i also corrected that