cpointerspointer-arithmeticformat-specifiers

Unexpected value when dereferencing pointer in C

The following code

#include <stdio.h>

int main()
{
    long long data = 0xFFFEABCD11112345;

    char *pData = (char *)&data;
    printf("Value at address %p is %x\n", pData, *pData);

    pData = pData + 5;
    printf("Value at address %p is %x\n", pData, *pData);

    return 0;
}

produces an output similar to

Value at address 00000023515FFC00 is 45
Value at address 00000023515FFC05 is ffffffab

Given that pData is a char *, I was expecting the second value to be ab instead of ffffffab. I believe that the %x format specifier might be the culprit but I do not fully understand it. Where do the leading f's come from?

Solution

  • char may be either signed or unsigned depending on compiler. Is char signed or unsigned by default? In this case it appears to be signed.

    On mainstream computers, signed char can only hold values -128 to 127. 0xAB on such a computer would be the 2's complement representation of a decimal negative number -85.

    C has various forms of implicit type promotions that happen when small types like char are used in most expressions, or as in this case when passed along to a variadic function printf. The special set of implicit promotion rules for variadic functions are called "the default argument promotions" and they state that small integer types get promoted to int regardless of signedness.

    In case we have a signed char with value -85, then during promotion to int the sign is respected, which is known as sign extension. Meaning the value is still -85 but the binary 2's complement representation of the promoted int may be 0xFFFFFFAB (assuming 32 bit int).

    However, if we had unsigned char with the value 0xAB/171, then during promotion to int the value is just kept and no sign is present. So we could have avoided sign extension by casting: (unsigned char)*pData. The explicit conversion from signed to unsigned is well-defined.

    The format string of printf holds no relevance for this promotion. %x expects a parameter which is unsigned int so we essetially lie to print since we pass a char promoted to int, strictly speaking undefined behavior. However, printf in this case just reads the binary representation of the int and presents it as 0xFFFFFFAB.

    Take away: