Bit right shifting of C language in macOS 10.12 is different from Ubuntu 16.04

Question

Update : I just rewrote the function in a new C source file on macOS:

#include <stdio.h>

int main() {

int x = 0xffffffff;

int m2 = (((((0x55 << 8) + 0x55 )<< 8) + 0x55)<< 8) + 0x55;
printf("m2 : 0x%x\n",m2);

int m4 = (((((0x33 << 8) + 0x33 )<< 8) + 0x33)<< 8) + 0x33;
printf("m4 : 0x%x\n",m4);

int m8 = (((((0x0f << 8) + 0x0f )<< 8) + 0x0f)<< 8) + 0x0f;
printf("m8 : 0x%x\n",m8);

int m16 = (0xff << 16) + 0xff ;
printf("m16 : 0x%x\n",m16);

int p1 = (x & m2) + ((x >> 1) & m2);
printf("p1: 0x%x\n",p1);

printf("p1 & m4 : 0x%x\n",p1 & m4);

printf("p1 >> 2 : 0x%x\n",p1 >> 2);

printf("(p1 >> 2) & m4 : 0x%x\n",(p1 >> 2) & m4);

int p2 = (p1 & m4) + ((p1 >> 2) & m4);
printf("p2 : 0x%x\n",p2);

int p3 = (p2 & m8) + ((p2 >> 4) & m8) ;
printf("p3 : 0x%x\n",p3);

int p4 = (p3 & m16)  + ((p3 >> 8) & m16) ;
printf("p4 : 0x%x\n",p4);
//int p4 = p3   + (p3 >> 8)  ;
int p5 = p4 + (p4 >> 16) ;

printf("BigCount result is : 0x%x\n",p5 & 0xFF);
}

And the printed result is :

everything is as same as the one in Ubuntu. This makes me more confused.

---

When I run this function in macOS 10.12, it gave an unexpected answer. The input x is 0xffffffff (-1). The function is written in C language:

int bitCount(int x) {

  int m2 = (((((0x55 << 8) + 0x55 )<< 8) + 0x55)<< 8) + 0x55;
  int m4 = (((((0x33 << 8) + 0x33 )<< 8) + 0x33)<< 8) + 0x33;
  int m8 = (((((0x0f << 8) + 0x0f )<< 8) + 0x0f)<< 8) + 0x0f;
  int m16 = (0xff << 16) + 0xff ;

  int p1 = (x & m2) + ((x >> 1) & m2);
  int p2 = (p1 & m4) + ((p1 >> 2) & m4); //line 7
  int p3 = (p2 & m8) + ((p2 >> 4) & m8) ;
  int p4 = (p3 & m16)  + ((p3 >> 8) & m16) ;
  //int p4 = p3   + (p3 >> 8)  ;
  int p5 = p4 + (p4 >> 16) ;

  return p5 & 0xFF;
}

When I track all the local variable by print it, I just found that the:

((p1 >> 2) & m4) //line7

printed the value of '0x2222222'(seven '2' instead of eight '2').

This is so unexpected, the p1 prints '0x2aaaaaaa' and the m4 is '0x33333333', so it should be 0x22222222(eight '2;). However, when I run this in ubuntu 16.04, everything is just as expected, for the ((p1 >> 2) & m4) prints '0x22222222':

Do you run this on your Mac having the same problem? Does anything different in macOS lead to this problem?

Answer 1

Per 6.5.7 Bitwise shift operators, of the C standard (note the highlighted parts):

Constraints

2 Each of the operands shall have integer type.

Semantics

3 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.

4 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 2E2 , 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 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

5 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 / 2E2 . If E1 has a signed type and a negative value, the resulting value is implementation-defined.

You are bit-shifting signed integer values. Depending on your inputs, you are depending on both undefined and implementation-defined behavior.

Different results on different platforms are to be expected.