In C bit-fields, predict the printed values when using 1-bit and 4-bit signed fields.\n\n#include <stdio.h>\n\nint main()\n{\n struct value {\n int bit1 : 1;\n int bit3 : 4;\n int bit4 : 4;\n } bit = { 1, 2, 13 };\n\n printf("%d, %d, %d\n", bit.bit1, bit.bit3, bit.bit4);\n return 0;\n}\n

Introduction / Context:
The task examines signed bit-field behavior, especially how values outside the representable range wrap and are interpreted as negative numbers for very small signed widths.

Given Data / Assumptions:

• bit1 is a signed 1-bit field, initialized with 1.
• bit3 is a signed 4-bit field, initialized with 2.
• bit4 is a signed 4-bit field, initialized with 13.
• printf prints the three fields in that order using %d.

Concept / Approach:
A 1-bit signed field has range -1 to 0. Storing 1 yields the pattern that represents -1. A 4-bit signed field has range -8 to 7. Storing 13 causes wrapping modulo 16, producing 13 - 16 = -3. The middle field stores 2, which is within range and remains 2.

Step-by-Step Solution:

Verification / Alternative check:
Consider two’s complement encoding: for 1-bit signed, the single bit 1 is the sign bit set, which is -1. For 4-bit signed, 1101 is -3, confirming the wrap logic.

Why Other Options Are Wrong:

• 1, 2, 13: Treats fields as wider unsigned; not correct for signed narrow fields.
• 1, 4, 4: Ignores signedness and the actual initializers.
• -1, -2, -13: Only the first and last could be negative; middle is 2, not -2.

Common Pitfalls:
Forgetting that small signed bit-fields have very tight ranges and that overflow wraps into negative representations. Also confusing signed and unsigned bit-fields leads to wrong conclusions.

Final Answer:
-1, 2, -3