Why Does `printf` Produce Unexpected Hexadecimal Output When Printing Characters?

Understanding printf Hexadecimal Printing Behavior

When printing a hexadecimal representation of a character using printf, unexpected results can arise if a char is passed without explicit casting. This is due to the behavior of printf, which expects an unsigned int parameter for the %x modifier.

Char Promotion and Undefined Behavior

In C , characters (char) are typically promoted to int within varargs functions such as printf. However, the size of an int varies between platforms, potentially causing undefined behavior if the char is not explicitly cast.

Explicit Casting for Predictable Results

To ensure predictable results, explicitly cast the char to an unsigned int before printing. This will prevent unexpected behavior and display the correct hexadecimal representation:

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

Field Width Considerations

Note that setting a field width of 1 for hexadecimal printing is not particularly useful, as it specifies the minimum number of digits to display and at least one digit is always necessary.

Unsigned Char Handling

In case char on the platform is signed, it's important to handle it appropriately. To avoid converting negative char values to large unsigned int values, consider using unsigned char or explicitly casting the value using unsigned char or a masking operation:

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

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

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