c - Extracting the sign bit with shift -

is defined behavior extract sign of 32 bit integer way:

#include <stdint.h>  int get_sign(int32_t x) {     return (x & 0x80000000) >> 31; } 

do result of 0 or 1?

no, incorrect because right shifting signed integer negative value implementation-defined, specified in c standard:

6.5.7 bitwise shift operators

the result of e1 >> e2 e1 right-shifted e2 bit positions. if e1 has unsigned type or if e1 has signed type , nonnegative value, value of result integral part of quotient of e1 / 2e2. if e1 has signed type , negative value, resulting value implementation-defined.

you should cast x (uint32_t) before masking , shifting.

edit: wrong answer! shall keep answer here example of looking, intuitive incorrect reasoning. explained in other answers, there not right shifting of negative value in code posted. type of x & 0x80000000 1 of signed integer or unsigned integer types depending on implementation characteristics, value positive, either 0 or 2147483648. right shifting value not implementation-defined, result either 0 or 1. whether result value of sign bit less obvious: value of sign bit except contorted corner cases, hybrid architectures quite unlikely exist , non standard conforming anyway.