Question

I’m getting results from left shift to which I could not find an explanation.

unsigned char value = 0xff; // 1111 1111
unsigned char  = 0x01; // 0000 0001

std::cout << "SIZEOF value " << sizeof(value) << "\n"; // prints 1 as expected
std::cout << "SIZEOF shift " << sizeof(shift) << "\n"; // prints 1 as expected

std::cout << "result " << (value << shift) << "\n"; // prints 510 ???

std::cout << "SIZEOF result " <<  sizeof(value << shift) << "\n"; // prints 4 ???

I was expecting result to be 1111 1110 but instead I get int(?) with value of 1 1111 1110.

How can the bits of an unsigned char be shifted to the left so that bits are truncated and the result is 1111 1110?

What I’m trying to do is to read series of bytes and interpret them as integers of varying lengths (1-32 bits).

F0        F5
1111 0000 1111 0101 

could be

0F (first 4 bits)
0F (next 8 bits)
05 (last 4 bits)

Has this something to do with the fact that arithmetic is not done with types smaller than int?

Answer 1

Quoting some draft of the 2011 standard:

5.8 Shift operators [expr.shift]

…

The operands shall be of integral or unscoped enumeration type and integral promotions are performed.
The type of the result is that of the promoted left operand.

and

4.5 Integral Promotions [conv.prom]

A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion
rank (4.13) is less than the rank of int can be converted to a prvalue of type int if int can represent all
the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned
int

…

So, value is promoted to int, and the type of value << shift is the type of the promoted left operand, i.e. int.

You can achieve your desired result one of these ways:

std::cout << "result " << ((unsigned char)(value << shift)) << "\n";
std::cout << "result " << ((value << shift)&0xff) << "\n";