Question

The question should be basic but i am surprised that i had some trouble get it now. First one is when i glanced ‘C++ primer’ book chapter 5.3. The Bitwise Operators, when author use below code as an example to explain shift operation:

unsigned char bits = 1;     // '10011011' is the corresponding bit pattern
bits << 1;                  // left shift

My head spin a little when i looked at this, where is ‘10011011’ coming from? ‘1’ is not ‘0x01’?

Another question comes from http://c-faq.com/strangeprob/ptralign.html, where author try to unpack structure:

struct mystruct {
    char c;
    long int i32;
    int i16;
} s;

using

unsigned char *p = buf;

s.c = *p++;

s.i32 = (long)*p++ << 24;
s.i32 |= (long)*p++ << 16;
s.i32 |= (unsigned)(*p++ << 8);  // this line !
s.i32 |= *p++;

s.i16 = *p++ << 8;
s.i16 |= *p++;

My question is, p is a pointer to unsigned char(which is 8 bits), right? when building higher bytes of s.i32(24~31, 16~23), *p++ is converted to ‘long'(32bits) before doing left shift, so left shift would not lose bit in *p++, but in

s.i32 |= (unsigned)(*p++ << 8);

*p++ is shifted first, then convert to unsigned int, wouldn’t the bits of *p++ all lost during the shift?

Again i realize i maybe missing some of the basics in C here. Hope someone can give a hand here.

Thanks,

Answer 1

To answer your second question, performing any arithmetic operation on a char promotes it to a (possibly unsigned) int (including shifting and other bitwise operations), so the integer-size value is shifted, not the 8-bit char value.