Question

1.
Consider the following:

unsigned int a, b;

b = a >> ((sizeof a) * CHAR_BIT);

/* or 2nd operand greater than ((sizeof a) * CHAR_BIT) */

Is this is defined, undefined behavior or implementation dependent behavior?

2.
Also another sub-question:

In the case a is signed int and it is shifted less than its bit length, is the signed bit shifting implementation defined or undefined behavior. In both the cases:

1. When shifting right : a >> 5
2. When shifting left : a << 5

EDIT question edited

Answer 1

1.

From C99 standard, section 6.5.7:

The integer promotions are performed on each of the operands. The type of the result is
that of the promoted left operand. If the value of the right operand is negative or is
greater than or equal to the width of the promoted left operand, the behavior is undefined.

So it’s undefined.

2.

From the same section:

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 x 2^E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 x 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2^E2. If E1 has a signed type and a negative value, the resulting value is implementation-defined.

So for left-shift, it’s well-defined if a is signed and positive. It’s undefined if a is signed and negative.

For right-shift, it’s well-defined if a is signed and positive. It’s implementation-defined if a is signed and negative.