Question

In C, on a 32-bit machine, I was just wondering if 1>>31 returns -1 given 1 is a signed integer, since for 2’s-complement, while doing right shift (arithmetic), sign bit is copied giving the result

1111 1111 1111 1111 1111 1111 1111 1111

Answer 1

No, the result will be zero in any conforming implementation.

C99, 6.5.7/5 (“Bitwise shift operators”) states:

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.

Since 1 is nonnegative, the result is the integral quotient of 1 / (2^31) which is obviously zero.