Bit shifts with negative counts in C: determine the safest conclusion about this program’s outputs (assume typical hosted C, int is 32 bits).\n\n#include<stdio.h>\n\nint main()\n{\n printf("%d %d\n", 32<<1, 32<<0);\n printf("%d %d\n", 32<<-1, 32<<-0);\n printf("%d %d\n", 32>>1, 32>>0);\n printf("%d %d\n", 32>>-1, 32>>-0);\n return 0;\n}\n\nNote: shifting by a negative amount is not defined by the C standard.

Introduction / Context:
This question gauges knowledge of defined vs. undefined behavior for bit shifts in C. While shifting by nonnegative counts is defined, any negative shift count leads to undefined behavior, meaning the program has no guaranteed result.

Given Data / Assumptions:

• 32<<1 and 32<<0 are well-defined: 64 and 32.
• 32>>1 and 32>>0 are well-defined: 16 and 32.
• 32<<-1 and 32>>-1 have negative shift counts → undefined behavior.
• -0 is just 0, so shifts by -0 behave like 0 and are defined.

Concept / Approach:
Per the C standard, shifting by a negative count is undefined. Compilers may produce any result, crash, or optimize strangely. Therefore, any option claiming specific numeric values for those lines is unreliable; the safe, standard-aware answer is that outputs are indeterminate for the lines with negative counts.

Step-by-Step Solution:
Line 1: prints 64 32.Line 2: first shift is undefined; second is 32.Line 3: prints 16 32.Line 4: first shift is undefined; second is 32.

Verification / Alternative check:
Different compilers or optimization levels may show different numbers or traps for negative shifts, proving the behavior is not portable.

Why Other Options Are Wrong:
Any fully specified numeric sequence asserts a defined outcome for the negative shifts, which is not portable or guaranteed.

Common Pitfalls:
Assuming negative shifts “wrap” or always return 0; writing code that relies on compiler-specific extensions.

Final Answer:
Garbage values