In C bit-fields, what is printed for a 1-bit signed field initialized to 1?\n\n#include <stdio.h>\n\nint main()\n{\n struct byte {\n int one : 1;\n };\n struct byte var = { 1 };\n printf("%d\n", var.one);\n return 0;\n}\n

Introduction / Context:
The problem focuses on how a signed 1-bit bit-field represents values and how printing that field behaves when it is initialized with the literal value 1.

Given Data / Assumptions:

• Field width is 1 bit and type is signed (since it is int).
• Initializer: one = 1.
• Output uses %d to print the value.

Concept / Approach:
A signed 1-bit quantity can represent only -1 and 0 in two’s complement. The bit pattern 1 is interpreted as -1, and the bit pattern 0 is 0. Therefore assigning 1 yields a stored representation that prints as -1 for a signed field.

Step-by-Step Solution:

Verification / Alternative check:
If the field were declared unsigned (e.g., unsigned int one : 1), then printing would show 1. The signedness is the key here.

Why Other Options Are Wrong:

• 1: Correct only for an unsigned 1-bit field.
• 0: Would require the stored bit to be 0.
• Compilation error: The code is valid C.

Common Pitfalls:
Forgetting that the base type of a bit-field controls signedness, and assuming that a 1 stored in a 1-bit field must print as 1 regardless of signedness.

Final Answer:
-1