C programming/ Predict the output

Question

main()
{
    {
        extern int i;
        int i=20;
        {
            const volatile unsigned i=30; printf("%d",i);
        }
        printf("%d",i);
    }
printf("%d",i);
}
int i;

Comments

but global variables are always intialized to 0. right?

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right...

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only for c language but not other language like java bcz they have garbage collector or some thing like that

Answer 1

I am sure you know what is declaration, what is definition and what is "extern". If you do not know then i will suggest this link for you, Read this and you will be through. 

Now Let me explain, the output just in your program. 

main()
{
    {
        extern int i;   // This line declare the variable i, but it doe not
                        // allocate memory location. It say that "i" is a variable
                        // of type int and its somewhere in the program.
        
        int i=20;   // This line is the definition for the above declaration. Now 
                    // Memory for "i" is allocated here and assigned the value 20
        {
            const volatile unsigned i=30; //Here "const" make variable i constant
                                        // volatile is another keyword used in the C
                                        // program which means that value of i can be
                                        // chenged by compiler for optimization purpose
            
            printf("%d",i); // This will print local i, which is 30
        }
        
        printf("%d",i); // This will print local i, here local i is 20
    }
    
    printf("%d",i); // This will access global i. hence print 0
}
int i; //Global i. since we have not assigned any value to this variable, compiler
        // assigned 0 to this i
        
// Hence you get output 30 20 0

See, Its a very simple and straight forward question. As I always say that whenever you see a question then ask yourself why this question. So for this question why is : Static scoping. extern storage class, default initial value of global variable, use of const and volatile, difference of declaration and definition. Hope this help. Cheers. 

Comments

Won't the last printf option print a Grabage Value instead of Zero as 'i'  is not declared as EXTERN ?

Answer 2

'{' introduces new block and thus new scope. In the innermost block i is declared as,
      const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i 

has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no

 storage space is allocated for it. After compilation is over the linker resolves it to global 

variable i (since it is the only variable visible there). So it prints i's value as 0. 

so answr will be 30200