Bitwise operations performed on an 8 bit unsigned int

Question

In the following code, bitwise operations are performed on an 8-bit unsigned integer.

uint8_t i = 10;
uint8_t j = i>>2;

In this example, i is promoted to a signed int and value assigned to 8 bits unsigned int. Is it safe to int converted back to 8 bits unsigned int?

I don't think it matters: assignment will truncate the result anyway. And with your values it's not an issue. but your question has a lot of typos. Can you edit it? — Jean-François Fabre, Nov 29 '17 at 16:57
And I don't really see what exactly are you asking here. Are you asking if it is guaranteed to result in `2`? Yes it is. — Eugene Sh., Nov 29 '17 at 16:59
*after the edit* - the conversion process is covered by the standard [here](http://port70.net/~nsz/c/c11/n1570.html#6.3.1.3) .It is safe to covert any integer to an unsigned integer type in terms it is always well defined. — Eugene Sh., Nov 29 '17 at 17:39

Petr Skocik · Accepted Answer · 2017-11-29T17:27:58.377

No.

In arithmetic expressions, the lesser types are converted to int if possible, and otherwise to unsigned int (see 6.3.1.1p2).

Practically,

uint8_t i = 10;
puts(_Generic(i>>2, uint8_t: "u8", int: "int", unsigned int: "uint"));

should output "int", as should

uint8_t i = 10, j = 2;
puts(_Generic(i>>j, uint8_t: "u8", int: "int", unsigned int: "uint"));

(In the former example, i would also get promoted to int because 2 is of type int and because usual arithmetic conversions would apply.)

Bitwise operations performed on an 8 bit unsigned int

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