Unsigned integer bit field shift yields signed integer

Question

Let consider the following program test.c:

#include 

struct test {
    unsigned int a:5;
};

int main () {
    unsigned int i;
    s         


        

Answer 1

Integer promotions are applied to the operands of shift covered in the draft C99 standard section 6.5.7 Bitwise shift operators paragraph 3 which says (emphasis mine going forward):

The integer promotions are performed on each of the operands.[...]

and the integer promotion of a bit-field is covered in section 6.3.1.1 Boolean, characters, and integers paragraph 2 which says:

The following may be used in an expression wherever an int or unsigned int may be used:

and contains the following bullet:

— A bit-field of type _Bool, int, signed int, or unsigned int.

and then says:

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.⁴⁸⁾ All other types are unchanged by the integer promotions.

it was clarified in the draft C11 standard:

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.⁵⁸⁾ All other types are unchanged by the integer promotions.

So this is expected behavior.