As an exercise I have to write the following function:
multiply x by 2, saturating to Tmin / Tmax if overflow, using only bit-wise and bit-shift operations.
Now this is my code:
// xor MSB and 2nd MSB. if diferent, we have an overflow and SHOULD get 0xFFFFFFFF. otherwise we get 0.
int overflowmask = ((x & 0x80000000) ^ ((x & 0x40000000)<<1)) >>31;
// ^ this arithmetic bit shift seems to be wrong
// this gets you Tmin if x < 0 or Tmax if x >= 0
int overflowreplace = ((x>>31)^0x7FFFFFFF);
// if overflow, return x*2, otherwise overflowreplace
return ((x<<1) & ~overflowmask)|(overflowreplace & overflowmask);
now when overflowmask
should be 0xFFFFFFFF
, it is 1 instead, which means that the arithmetic bit shift >>31
shifted in 0s instead of 1s (MSB got XORed to 1, then shifted to the bottom).
x is signed and the MSB is 1, so according to C99 an arithmetic right shift should fill in 1s. What am I missing?
EDIT: I just guessed that this code isn't correct. To detect an overflow it suffices for the 2nd MSB to be 1.
However, I still wonder why the bit shift filled in 0s.
EDIT:
Example: x = 0xA0000000
x & 0x80000000 = 0x80000000
x & 0x40000000 = 0
XOR => 0x80000000
>>31 => 0x00000001
EDIT:
Solution:
int msb = x & 0x80000000;
int msb2 = (x & 0x40000000) <<1;
int overflowmask = (msb2 | (msb^msb2)) >>31;
int overflowreplace = (x >>31) ^ 0x7FFFFFFF;
return ((x<<1) & ~overflowmask) | (overflowreplace & overflowmask);
0x80000000
is treated as an unsigned number which causes everything to be converted to unsigned, You can do this:
// xor MSB and 2nd MSB. if diferent, we have an overflow and SHOULD get 0xFFFFFFFF. otherwise we get 0.
int overflowmask = ((x & (0x40000000 << 1)) ^ ((x & 0x40000000)<<1)) >>31;
// this gets you Tmin if x < 0 or Tmax if x >= 0
int overflowreplace = ((x>>31)^0x7FFFFFFF);
// if overflow, return x*2, otherwise overflowreplace
return ((x<<1) & ~overflowmask)|(overflowreplace & overflowmask);
OR write the constants in negative decimals
OR I would store all the constants in const int variables to have them guaranteed signed.