What Does It Mean That "A Declaration Shadows a Parameter"?
I Am Trying to Make a Function That Returns Double the Integer Number That I Will Pass to It. I Am Getting the Following Error Message with My Code...
I am trying to make a function that returns double the integer number that I will pass to it. I am getting the following error message with my code:
declaration of 'int x' shadows a parameter int x; "
Here is my code:
#include <iostream>
int doublenumber();
using namespace std;
int doublenumber(int x)// <-- this is the function which returns double the value .
{
int x;
return 2 * x;
cout << endl;
}
int main()
{
int a;
cout << "Enter the number that you want to double it : " << endl;
cin >> a;
doublenumber(a);
return 0;
}
3 Answers
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.
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;
}
#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.
Sarah Jenkins is a veteran tech journalist with over 12 years of experience covering artificial intelligence, mobile innovations, and digital ethics. Her insights have appeared in leading technology publications worldwide.
