Operator "<<= " : What does it it mean?

Question

I need help solving this problem in my mind, so if anyone had a similar problem it would help me.

Here's my code:

char c=0xAB;
printf("01:%x\n", c<<2);
printf("02:%x\n", c<<=2);
printf("03:%x\n", c<<=2);

Why the program prints:

01:fffffeac
02:ffffffac
03:ffffffb0

What I expected to print, that is, what I got on paper is:

01:fffffeac
02:fffffeac
03:fffffab0

I obviously realized I didn't know what the operator <<= was doing, I thought c = c << 2 .

If anyone can clarify this, I would be grateful.

Answer 1

You're correct in thinking that

c <<= 2

is equivalent to

c = c << 2

But you have to remember that c is a single byte (on almost all systems), it can only contain eight bits, while a value like 0xeac requires 12 bits.

When the value 0xeac is assigned back to c then the value will be truncated and the top bits will simply be ignored, leaving you with 0xac (which when promoted to an int becomes 0xffffffac ).

Answer 2

<<= means shift and assign. It's the compound assignment version of c = c << 2; .

There's several problems here:

• char c=0xAB; is not guaranteed to give a positive result, since char could be an 8 bit signed type. See Is char signed or unsigned by default? . In which case 0xAB will get translated to a negative number in an implementation-defined way. Avoid this bug by always using uint8_t when dealing with raw binary bytes.

• c<<2 is subject to Implicit type promotion rules - specifically c will get promoted to a signed int . If the previous issue occured where your char got a negative value, c now holds a negative int .

• Left-shifting negative values in C invokes undefined behavior - it is always a bug. Shifting signed operands in general is almost never correct.

• %x isn't a suitable format specifier to print the int you ended up with, nor is it suitable for char .

As for how to fix the code, it depends on what you wish to achieve. It's recommended to cast to uint32 before shifting.