C++ defaults, compound assignments, and parameter passing by value: trace member updates and print x and y. #include <iostream.h> class CuriousTab { int x, y, z; public: void Apply(int xx = 12, int yy = 21, int zz = 9) { x = xx; // x = 12 y = yy += 10; // yy becomes 10 locally; y = 10 z = x -= 2; // x becomes 10; z = 10 } void Display() { cout << x << " " << y << endl; } void SetValue(int xx, int yy) { Apply(xx, 0, yy); // pass-by-value } }; int main() { CuriousTab *pCuriousTab = new CuriousTab; (*pCuriousTab).SetValue(12, 20); pCuriousTab->Display(); delete pCuriousTab; return 0; }

Introduction / Context: This exercise tests understanding of default arguments, pass-by-value, and chained assignments with compound operators. It asks you to follow exact state changes inside Apply and then print two members using Display.

Given Data / Assumptions:

• SetValue calls Apply(12, 0, 20).
• Parameters are passed by value; modifications to yy do not escape Apply except through explicit assignments to members.
• Operators used: yy += 10 and x -= 2, with their results assigned to y and z respectively.

Concept / Approach: Carefully evaluate compound assignments which both change the left-hand operand and yield a value. Keep track of x and y after each statement to determine what Display prints.

Step-by-Step Solution:

Verification / Alternative check: Rewriting as yy = yy + 10; y = yy; x = x - 2; z = x; confirms the same final member values.

Why Other Options Are Wrong:

• 12 10: Ignores the later x -= 2.
• 12 21 / 12 31: Misapply defaults or assume pass-by-reference, which is not used here.
• The program will report compilation error.: The code compiles as intended.

Common Pitfalls: Thinking yy += 10 returns the old value, or assuming parameters are references; both lead to incorrect y values.

Final Answer: 10 10