Why Does `printf` Show Unexpected Hex Output When Printing a Byte?

Understanding printf's Unexpected Output When Printing a Byte as Hex

When working with vector data types like pixel_data containing char elements, it is common to encounter unexpected results when attempting to print a single byte in hexadecimal format. Specifically, printing pixel_data[0] using printf(" 0x%1x ", pixel_data[0] ) might yield 0xfffffff5 instead of the anticipated 0xf5.

This behavior arises from a characteristic of printf, where the %x modifier anticipates an unsigned integer as its argument. As a result, the char value, which is typically promoted to an integer when supplied to a varargs function, is treated as an int by printf.

To obtain dependable results, an explicit cast to an unsigned int is recommended:

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

Additionally, a field width of one may not be ideal, as it specifies a minimum number of digits to display, and at least one digit will always be used.

If char on your system is handled as a signed type, the conversion may result in the generation of large unsigned int values (e.g., fffffff5) for negative char values. To ensure byte values are treated as unsigned, use unsigned char for pixel_data, cast via unsigned char, or employ a masking operation:

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

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