What does it mean that “a declaration shadows a parameter”? [closed]

Answer 1

You have x as a parameter and then try to declare it also as a local variable, which is what the complaint about "shadowing" refers to.

Answer 2

I did it because your advice was so helpful, and this is the final result :

#include <iostream>
using namespace std;

int doublenumber(int x)
{
    return 2*x;
}

int main()
{
    int a;
    cout << "Enter the number that you want to double it : " << endl;
    cin>>a;
    int n= doublenumber(a);
    cout << "the double value is : " << n << endl;
    return 0;
}

Answer 3

#include <iostream>
using namespace std;
int doublenumber(int x)
{
return 2*x;
}
int main()
{
int a;
cout << "Enter the number that you want to double it : " << endl;
cin>>a;
int d = doublenumber(a);

cout << "Double : " << d << endl;

return 0;
}

There are some problem with your code. Your declaration and definition of function dies not match. So remove declaration as no necessity of it.

You are declaring local x variable inside function which will shadow your function arguments.