Bit shifting and creating a mask

Question

I am creating a method that takes 3 integers (start, end, z), the result should be a mask of either 1 or 0 depending on z from start to end.

For example, if getMask(2,6,1) is called the result is: 00000000000000000000000001111100

For some reason when I call getMask(0,31,1) I get: 00000000000000000000000000000000 instead of 11111111111111111111111111111111

My method code is:

if (z == 1) { return ~(~0 << (end - start + 1)) << (start); } else if (z == 0) { return ~(~(~0 << (end - start + 1)) << (start)); }

I want to know why I would be getting that result instead of the expected one.

Edit: So i understand why it is happening but how do I fix it?

You're using int, which is likely 32 bits on your system. Unsigned shifting is performed modulo number of bits, but signed shifts (what you're doing) are undefined behavior if you shift >= the number of total bits. Reference — Cruz Jean
– Cruz Jean, Commented Apr 7, 2019 at 19:58

padawin · Accepted Answer · 2019-04-07 21:00:39Z

Here is a working version to it:

#include "limits.h" #include <stdio.h> unsigned int get_mask(int start, int end, int z); void p(unsigned int n); int main() { p(get_mask(2, 6, 1)); p(get_mask(0, 31, 1)); p(get_mask(0, 31, 0)); p(get_mask(1, 31, 1)); p(get_mask(1, 31, 0)); p(get_mask(6, 34, 1)); } unsigned int get_mask(int start, int end, int z) { int rightMostBit = end - start + 1; unsigned int res = ~0; if (rightMostBit < sizeof(int) * CHAR_BIT) { res = ~(res << rightMostBit); } res = res << start; if (z == 0) { res = ~res; } return res; } void p(unsigned int n) { for (int i = 31; i >= 0; --i) { if (n & (1 << i)) { printf("1"); } else { printf("0"); } } printf("\n"); }

I basically do the first shift and ~ only if it does not overflow. If it does, the initial ~0 is already ready to be shifted of start bits.

Would this work if it returned a regular int and not unsigned. That is a requirement for my assignment.
yes it would. I invite you to test it and play a bit with it.

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