C++ constructor pre-increment parameters and post-increment in output: what does Show() print?\n\n#include<iostream.h>\nclass CuriousTabData\n{\n int x, y, z;\npublic:\n CuriousTabData(int xx, int yy, int zz)\n {\n x = ++xx; y = ++yy; z = ++zz;\n }\n void Show()\n {\n cout << "" << x++ << " " << y++ << " " << z++;\n }\n};\nint main()\n{\n CuriousTabData objData(1, 2, 3);\n objData.Show();\n return 0;\n}

Introduction / Context:
This question checks the effect of pre-increment on constructor parameters versus post-increment when streaming values. Understanding sequence and timing of increments is key.

Given Data / Assumptions:

• Constructor arguments are (1,2,3).
• Assignments use pre-increment: x=++xx, y=++yy, z=++zz.
• Printing uses post-increment: x++, y++, z++.

Concept / Approach:
Pre-increment increments first, then yields the incremented value for assignment. Post-increment yields the current value for output, then increments afterward.

Step-by-Step Solution:
After construction: x=2, y=3, z=4.Streaming with post-increment prints 2, 3, 4 in order.After printing, the internal values become 3, 4, 5 (not shown).

Verification / Alternative check:
Add a second call to Show(); it will then print 3 4 5 because of the prior post-increments.

Why Other Options Are Wrong:
Option A ignores the constructor’s pre-increment. Option C reflects the internal state after a hypothetical second print. Option D is incorrect because the code is valid C++.

Common Pitfalls:
Mixing up pre- and post-increment semantics and assuming both behave identically in assignments and outputs.

Final Answer:
The program will print the output 2 3 4 .