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I have the following c code:

int foo = 0;
printf("output1: %08x\n", (~0x0) << (~0));
printf("output2: %08x", (~0x0) << (~foo));

which prints out: 

output1: ffffffff
output2: 80000000

Why does shifting by the same number produce a different result?

Toby
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    You are shifting *signed* values. By *signed* amounts. IIRC the latter is *undefined behavior*. – Some programmer dude Feb 08 '18 at 07:30
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    Both of them are undefined behaviour; the results you get are arbitrary. You can only shift by 0 to one less than the number of bits in the type. See [C11 §6.7.5 Bitwise shift operators](https://port70.net/~nsz/c/c11/n1570.html#6.5.7). – Jonathan Leffler Feb 08 '18 at 07:30
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    @Someprogrammerdude: a signed shift isn't specifically a problem; a negative shift is, and `~0` is negative (usually treated as `-1`). – Jonathan Leffler Feb 08 '18 at 07:33

2 Answers2

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Basically your snippet is an orgy in undefined behavior.

  • 0x0 literals as well as variables of type int are signed, meaning they can get negative values. ~0 is always a negative value.
  • Left-shifting a signed int with negative value invokes undefined behavior. (6.5.7/4)
  • Attempting to shift more positions than there's room in the left operand invokes undefined behavior. (6.5.7/3)
  • Attempting to shift a negative amount of bits invokes undefined behavior. (6.5.7/3)

Therefore anything can happen in this program, including the whole thing crashing and burning. As a rule of thumb, never use signed variables together with bitwise operators.

Lundin
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2

Some compiler may told you the reason:

warning: shift count is negative [-Wshift-count-negative]
printf("output1: %08x\n", (~0x0) << (~0));
                                 ^  ~~~~

ints are signed and complementing them may produce negative values, and shifting by negative amount is undefined.

For example on my machine it produces:

output1: e785ba48
output2: 80000000
Jean-Baptiste Yunès
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