In C++ (default pointer argument bound to a static global), what is printed? Note that a local variable named b shadows the global, but the call passes both pointers explicitly. #include<iostream.h> static int b = 0; void DisplayData(int *x, int *y = &b) { cout << *x << " " << *y; } int main() { int a = 10, b = 20; DisplayData(&a, &b); return 0; }

Introduction / Context:
This checks understanding of default arguments captured at compile time, name shadowing between a file-scope static and a local variable, and what happens when the caller supplies both parameters explicitly anyway.

Given Data / Assumptions:

• Global (file-scope) static int b = 0 initializes the default for y as &b.
• main declares a local b = 20 which shadows the global name in that scope.
• Call site explicitly passes &a and &b (the local one).

Concept / Approach:
Because the call provides y, the default is not used. Therefore, dereferencing *x yields 10 and *y yields 20 from the local b. The existence of the global static only matters if the second argument were omitted.

Step-by-Step Solution:

Verification / Alternative check:
If you called DisplayData(&a) with no second argument, the output would be "10 0" due to the default binding to the file-scope static b.

Why Other Options Are Wrong:

• 10 0: Would occur only when omitting the second argument.
• Garbage / compile error: The code is valid and pointers are initialized.

Common Pitfalls:
Confusing which b is referenced when both a global and local exist; explicit argument selection wins here.

Final Answer:
The program will print the output 10 20.