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A program that sets "char c = 200;" and prints c with %d in printf printed −56 when compiled and run with gcc for x86-64 Linux, and 200 with gcc for Arm Linux. Which of the following is the reason for the difference?

  1. AArm is big-endian, so the bytes are read in reverse order
  2. BWhether a char without signed or unsigned is signed or unsigned is decided by the implementation; it is signed on x86-64 and unsigned on Arm
  3. Cchar on x86-64 is 16 bits, and its upper bit is treated as the sign
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Answer: B. Whether a char without signed or unsigned is signed or unsigned is decided by the implementation; it is signed on x86-64 and unsigned on Arm

Whether a char without signed or unsigned is signed or unsigned is not fixed by the standard but left to the implementation (implementation-defined). On x86-64 Linux, char is signed and can only represent −128 to 127, so storing 200 gives 200 − 256 = −56. On Arm and RISC-V Linux, char is unsigned and 200 is stored as is. When handling byte data, use unsigned char or uint8_t to make the signedness explicit.

  • A. Arm Linux is generally little-endian, and byte order does not affect the value of a 1-byte char.
  • B. Correct. The signedness of char is implementation-defined: signed on x86-64 and unsigned on Arm Linux.
  • C. char is 1 byte in every implementation, and 8 bits on x86-64.
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