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I have the below code and I can't understand why the last line doesn't return 77594624. Can anyone help me write the inverse bitwise operation to go from 77594624 to 4 and back to 77594624?

 Console.WriteLine(77594624); Console.WriteLine((77594624 >> 24) & 0x1F); Console.WriteLine((4 & 0x1F) << 24);
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    Once you right shift all the bits away, there's no way to recover them. Commented Jul 18, 2020 at 21:51
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    You should do "circular shift" check it here: stackoverflow.com/a/185743/168941 Commented Jul 18, 2020 at 21:55
  • There is no way to invert this. For example, (83886079 >> 24) & 0x1F also equals 4, and so does (77594625 >> 24) & 0x1F. In those examples, I replaced 77594624 with a different number, but the operation had the same result of 4. If the operation was invertible, only one value of x would map f(x) = (x >> 24) & 0x1F to 4. Commented Jul 18, 2020 at 23:13

2 Answers 2

Reset to default
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When you bit shift a value you might "lose" bits during that operation. If you right shift the value 16, which is 0b10000 in binary, by 4, you will get 1.

0b10000 = 16 0b00001 = 1

But this is also the case for other numbers like 28, which is 0b11100.

0b11100 = 28 = 16 + 8 + 4 0b00001 = 1

So with the start point 1 you cannot go "back" to the original number by left shifting again, as there is not enough information/bits available, you don't know if you want to go back to 16 or 28.

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77594624 looks like this in binary, and the x's mark the part that is extracted by the right shift and bitwise AND:

000001000101000000000000000000000 xxxxx

Clearly some information is lost.

If the other parts of the number were available as well, then they could be reassembled.

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