Question

What is the output?
void func (int *b) {
*b = 1:
}
int main()
int *a;
int n;
a=&n;
*a=0;
func(a);
std cout << *a << std:: endi
return 0;
}​

Answer 1

Answer:

1

Explanation:

because the operator overloading is not valid until the array function gets exceeded