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cprintfspecifier

what happens when %x read signed char?

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i thought *(p3 + 3) will print 90 but it shows ffffff90 why does it happend? i guess MSB is 1, and %x is for reading unsinged hexadecimal integer so it reads 90 like minus integer but it is not clear and i cant find about this problem at printf reference https://cplusplus.com/reference/cstdio/printf/ is there anyone who explain this?

Solution

  • Use an unsigned char *.

    In your environment,

    • char is signed.
    • char is 8 bits.
    • Signed numbers use two's complement.

    So you have a char with a bit pattern of 9016. In this environment, that's -112. So you are effectively doing the following:

    printf( "%x", (char)-112 );

    When passing to variadric function like printf, the smaller integer types are implicitly promoted to int or unsigned int. So what's really happening is this:

    printf( "%x", (int)(char)-112 );

    So you're passing the int value -112. On your machine, that has the bit pattern FF FF FF 9016 (in some unknown byte order). %x expects an unsigned integer, and thus prints that bit pattern as-is.