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uint8_t r = 0x5A;
r |= 0x01;
r &= ~0x10;
r ^= 0x0F;
printf("0x%02X\n", r);

uint8_t is an 8-bit unsigned integer type. Which of the following is the output when the code above is executed?

  1. A0x44
  2. B0x4F
  3. C0x54
Show answer and explanation

Answer: A. 0x44

0x5A is 0101 1010 in binary. |= 0x01 sets bit 0, giving 0101 1011 (0x5B), and &= ~0x10 clears bit 4, giving 0100 1011 (0x4B). The final ^= 0x0F inverts the lower 4 bits, giving 0100 0100, that is, 0x44. | sets bits, & combined with ~ clears bits, and ^ inverts bits.

  • A. Correct. The value changes 0x5A → 0x5B → 0x4B → 0x44.
  • B. This is the value obtained by mistaking the last ^= for |= and setting the lower 4 bits (0x4B | 0x0F = 0x4F).
  • C. This is the value obtained by skipping &= ~0x10 and leaving bit 4 set (0x5B ^ 0x0F = 0x54).
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