What will be the output of the following C++ code?
#include <iostream>
using namespace std;
int fun(int=0, int = 0);
int main()
{
cout << fun(5);
return 0;
}
int fun(int x, int y) { return (x+y); }
a) -5
b) 0
c) 10
d) 5
Answer: d
Explanation: C++ allows to define such prototype of the function in which you are not required to give variable names only the default values. While in function definition you can provide the variable names corresponding to each parameter.
Related Posts
What is the other name of the macro?
What will be the output of the following C++ code?
#include <iostream>
using namespace std;
#define MAX 10
int main()
{
int num;
num = ++MAX;
cout << num;
return 0;
}What will be the output of the following C++ code?
#include <iostream>
using namespace std;
#define PR(id) cout << “The value of ” #id ” is “<< id
int main()
{
int i = 10;
PR(i);
return 0;
}What will be the output of the following C++ code?
#include <iostream>
using namespace std;
#define SquareOf(x) x * x
int main()
{
int x;
cout << SquareOf(x + 4);
return 0;
}What is the mandatory preprocessor directive for c++?
How many types of macros are there in c++?
Which symbol is used to declare the preprocessor directives?
Join The Discussion