Why Does `printf` Show Unexpected Output When Printing Hexadecimal Bytes from a Char Vector?

Unexpected Output from printf When Printing Hexadecimal Bytes

When working with a vector of chars (pixel_data), printing a single byte as a hexadecimal value (printf(" 0x%1x ", pixel_data[0])) may unexpectedly produce a four-byte integer (0xfffffff5) instead of the intended value (0xf5).

Cause of the Issue

Printf typically expects an unsigned integer parameter for the %x modifier. However, a char is promoted to an int when passed to a varargs function like printf. This promotion results in the printing of additional bytes.

Resolving the Issue

To ensure predictable results, explicitly cast the char to an unsigned int:

printf(" 0x%1x ", (unsigned)pixel_data[0]);

Considerations for Variable Types

• Define pixel_data as unsigned char to treat the bytes as unsigned values.
• Alternatively, cast to unsigned char within the printf statement:

printf(" 0x%x ", (unsigned)(unsigned char)pixel_data[0]);

• Use a masking operation to zero-extend when converting to unsigned int:

printf(" 0x%x ", (unsigned)pixel_data[0] & 0xffU);

Understanding Field Width

The %1x field width specifies the minimum number of digits to display. However, it has limited usefulness in this context as at least one digit is always needed.

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