C++ global and local variables

问题:

I have encountered the following code:

#include<iostream>
using namespace std;
int i = 1;
int main(int argc,char ** argv)
{
    int i = i;
    cout<<i<<endl; // which i?
    return 0;
}

It can pass the compile, but gives the wrong answer, how to explain this?

回答1:

the Local variable is accessible, analogous to calling two people with a same name, one inside the room, and one outside the room. The one who is in the scope you are trying to access it , hears it.

回答2:

The int i = i; statement in main() declares a local variable that hides the global variable.

It initializes itself with itself (which has an indeterminate value). So the global i simply isn't used.

回答3:

Variables in deeper scopes will override the variables with the same name in a higher scope. To access a global variable, precede the name with ::

回答4:

Variables in the innermost scope will override variables with the same Name without warning.

回答5:

int main()
{
    int i=i;
    return 0;
}

is correct.

So in your program, the global i is ignored when the local i is encountered and it is initialized to itself. You'll get a garbage value as a result.

回答6:

When you have two variable with same name, one is global and other is local. Then in this case local variable will get in use only in that particular scope. And global variable is unused.

Now, coming to your problem

#include<iostream>
using namespace std;
int i = 1;//Global Variable decleration
int main(int argc,char ** argv)
{
    int i = i; // Local to main
    cout<<i<<endl; // which i?
    return 0;
}

int i = i; compiles without any error but when you run the program it will produce error because local i has indeterminate value.

回答7:

In C++, it is possible to access the global variable if you have a local variable with the same name but you have to use the scope resolution operator ::

change the line:

int i = i;

to

int i = ::i;

and the program will compile and work

回答8:

The concept most answers are describing is called shadowing.