以十六进制打印数据结构值

Question

typedef struct AbcStruct{

   short        LD;
   short        EL;
   short        CL;        
   AbcStruct( short b, short res = 0, short lr = 1000): LD( b ), EL(res), CL( lr ) { }
};

int main () 
{

    struct AbcStruct A2(200, 100, 100);

    char *string_ptr = (char *)&A2;
    kk = sizeof(AbcStruct);

    while(kk--)
        printf(" %x ", *string_ptr++);
}

Output (AbcStruct in Hex):

ffffffc8  0  64  0  64  0

I'm wondering why the output of first element contains 4 bytes: ffffffc8 when I was expecting it wouldd just print c8 .

Thanks

Answer 1

It does this because the first bit of the char you're printing is 1 , thus the char is interpreted as a negative number. When you're printing using the default %x format, the value to be printed is interpreted as being an int (much larger than the char ). Thus, the sign gets copied to all other positions, making you see those f s in the output.

One fix would be to print using %hhx (you're telling printf that you're printing unsigned char values).

printf(" %hhx ", *string_ptr++);

Another fix will be to change the type of the string_ptr to be unsigned char

unsigned char *string_ptr = (char *)&A2;

Or, you can combine them.

Answer 2

You are telling printf() to treat what is returned from the address pointed to by string_ptr as an unsigned int (which I guess is 32-bits long on your system) rather than an unsigned char .

Try this:

unsigned char *string_ptr = (unsigned char *)&A2;
kk = sizeof(AbcStruct);
while(kk--)
{
    unsigned char c = *string_ptr++;
    printf(" %x ", (unsigned)c);
}

Answer 3

Use the width option in printf format specifier to print as many characters of the address as you would prefer. For instance, here you could use %2X as your format specifier.