Question

Well, this code was taken from a practice exam for a programming thing that i’m very under qualified for, but i’d still like to know the reason behind the output for this.

    #include <iostream>
    using namespace std;

    int main(){

    char g = 'X', t = '1', *k_ = &t;

    g = 'y';
    cout << *k_ << *k_ << endl;
    g = 'Y';
    cout << *k_;
    k_ = &g;
    t = 'Z';
    cout << *k_ << *k_ << *k_ << endl;

    return 0;

    }

I don’t understand why the output for the last cout is 1YYY if there are only three k_ pointers, and there’s that 1 that’s bothering me too.

Answer 1

#include <iostream>
using namespace std;

int main(){

char g = 'X', t = '1', *k_ = &t;

g = 'y';
cout << *k_ << *k_ << endl; // prints out 11 with a newline
g = 'Y';
cout << *k_; // prints out 1 without a newline, so next cout will be on the same line
k_ = &g; // k_ now holds g's address (remember g is now 'Y')
t = 'Z';
cout << *k_ << *k_ << *k_ << endl; // prints out YYY on the same line as 1 from above

return 0;

}