C on 16-bit DOS (Turbo C): what will this program print, and why does the cast chain not matter?\n\n#include<stdio.h>\n\ndouble i;\n\nint main()\n{\n (int)(float)(char) i; // no effect on i or sizeof\n printf("%d", sizeof(i));\n return 0;\n}

Introduction / Context:
This question tests understanding of C's sizeof operator and the effect (or lack of effect) of cast expressions that are not used. In classic 16-bit DOS compilers like Turbo C, the size of fundamental types can differ from today’s 32/64-bit compilers. Here, we focus on what actually gets printed and why the cast chain does not change the result.

Given Data / Assumptions:

• Environment: Turbo C on DOS (16-bit).
• Global variable: double i;
• Prints: printf("%d", sizeof(i));
• There is a cast chain (int)(float)(char) i; whose value is not used.

Concept / Approach:
The sizeof operator returns the size, in bytes, of its operand’s type at compile time. Unused cast expressions are evaluated only for their value category/type (and may be optimized away); they do not change the type or size of the declared object. On Turbo C, sizeof(double) is typically 8 bytes.

Step-by-Step Solution:
1) Identify the printed expression: sizeof(i) equals sizeof(double). 2) On Turbo C (16-bit), sizeof(double) is 8. 3) The cast chain forms a temporary value and is discarded; it does not alter i. 4) Therefore, the program prints 8.

Verification / Alternative check:
Remove the cast line entirely; the output remains the same because sizeof(i) depends only on i’s type, not on prior unused expressions.

Why Other Options Are Wrong:
4: On Turbo C, double is not 4 bytes.
16/22: These sizes do not match Turbo C's double.
0: sizeof never returns zero for an object type.

Common Pitfalls:
Assuming sizeof(double) is always the host’s modern 8/16 bytes; conflating type conversions with object size; believing that an unused cast changes the variable’s type.

Final Answer:
8