In C bit-fields, predict the printed values when using 1-bit and 4-bit signed fields.
#include
int main()
{
struct value {
int bit1 : 1;
int bit3 : 4;
int bit4 : 4;
} bit = { 1, 2, 13 };
printf("%d, %d, %d
", bit.bit1, bit.bit3, bit.bit4);
return 0;
}
-
A1, 2, 13
-
B1, 4, 4
-
C-1, 2, -3
-
D-1, -2, -13
Answer
Correct Answer: -1, 2, -3
Explanation
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:
bit1 (1-bit signed): possible values are -1, 0 → assigning 1 gives -1.bit3 (4-bit signed): range -8..7 → assigning 2 stays 2.bit4 (4-bit signed): 13 exceeds 7 → wrap with modulus 16 → 13 - 16 = -3.printf prints: -1, 2, -3.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