Global and local variables in C++ - c++

Question

I'm fairly new to C++, and I'm experiencing some issues when printing local and global variables. Consider this simple piece of code:
#include <cstdlib>
#include <iostream>
using namespace std;
/*
*
*/
int x = 10; // This x is global
int main() {
int n;
{ // The pair of braces introduces a scope
int m = 10; // The variable m is only accessible within the scope
cout << m << "\n";
int x = 25; // This local variable x hides the global x
int y = ::x; // y = 10, ::x refers to the global variable x
cout << "Global: " << y << "\n" << "Local: " << x << "\n";
{
int z = x; // z = 25; x is taken from the previous scope
int x = 28; // it hides local variable x = 25 above
int t = ::x; // t = 10, ::x refers to the global variable x
cout << "(in scope, before assignment) t = " << t << "\n";
t = x; // t = 38, treated as a local variableout was not declared in this scope
cout << "This is another hidden scope! \n";
cout << "z = " << z << "\n";
cout << "x = " << x << "\n";
cout << "(in scope, after re assignment) t = " << t << "\n";
}
int z = x; // z = 25, has the same scope as y
cout << "Same scope of y. Look at code! z = " << z;
}
//i = m; // Gives an error, since m is only accessible WITHIN the scope
int m = 20; // This is OK, since it defines a NEW VARIABLE m
cout << m;
return 0;
}
My goal is practicing the accessibility of the variables within the various scopes, and then printing them. However, I'm not able to figure out why when I try to print the last variable z, NetBeans gives me back the output 2025.
Here it follows my sample output:
10
Global: 10
Local: 25
(in scope, before assignment) t = 10
This is another hidden scope!
z = 25
x = 28
(in scope, after re assignment) t = 28
Same scope of y. Look at code! z = 2520
RUN FINISHED; exit value 0; real time: 0ms; user: 0ms; system: 0ms
Hope that somebody can help me understanding what is going on! :)

Answer 1

is not that z is holding the value 2520 is the fact that you ommit to add a new line operator between printing z and printing m...
you are doing:
cout << "Same scope of y. Look at code! z = " << z;
}
int m = 20;
cout << m;
but you should do:
std::cout << "Same scope of y. Look at code! z = " << z << std::endl;
}
int m = 20;
std::cout << m << std::endl;
if you just followed the same criteria of labeling the output and doing something like
std::cout << "M is: "<<m << std::endl;
you would spotted the issue faster by observing the output:
25M is: 20