The P-Bug
From Andrey
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| Unfortunately, the retrieved value, as shown by debugger, is not '''130''', but '''-126'''. | Unfortunately, the retrieved value, as shown by debugger, is not '''130''', but '''-126'''. | ||
| - | If you think that this is the bug I'm talking about, it is not. This is just a usual signed/unsigned issue. Since c is declared as a (signed) <code>char</code>, the moment the 1 hits its upper bit it becomes a negative value. This is not a bug - this is how it supposed to be. The bits are all the same. It is still '''130''' under the hood. Let's move on. | + | If you think that this is the bug I'm talking about, it is not. This is just a usual signed/unsigned issue. Since c is declared as a (signed) <code>char</code>, the moment the 1 hits its upper bit it becomes a negative value. This is not a bug - this is how it supposed to be. The bits are all the same. It is still '''130''' under the hood. Let's continue. |
| Retrieve the same value as an integer: | Retrieve the same value as an integer: | ||
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| </pre> | </pre> | ||
| - | Which message will we get? On the one hand, this is a case of '''-126''' vs. '''130'''. On the other: | + | Which message will we get? On the one hand, this is a case of '''-126''' vs. '''130'''. On the other, we know that: |
| - | - '''Both values have same bits''' - 10000010, read from the same database field. | + | - '''Both values have same bits''' - 10000010, which are read from the same database field. |
| - | - '''Both are signed values''', so there is no signed/unsigned mismatch here. | + | - '''Both values are signed values''', so there is no signed/unsigned mismatch here. |
| - | - When evaluating the expression, both values are converted to same type, '''int'''. So when compared, they are of '''same size'''. | + | - When evaluating the expression, both values are converted to same type, '''int'''. So when compared, they are of '''same size'''. In other words, compiler sees <code>if ((int)c == i)<code>. |
| Which message will you see? I have to tell you. Despite of these assumptions, you'll be getting the else branch. This is what we call Potekhin's bug. | Which message will you see? I have to tell you. Despite of these assumptions, you'll be getting the else branch. This is what we call Potekhin's bug. | ||
Revision as of 02:12, 4 February 2006
Dear colleague,
Once in a while people put their names on things that they discover. We have Rorschach spots, Alzheimer's disease and Eiffel tower. Let me introduce Potekhin's Bug.
Let's put this value into database:
130
Let's retrieve it into a char using a usual ADO call:
char c = (char)(short)f->Item["MyField"]->Value;
In the above code, we need to cast to short since ADO doesn't know how to cast to char directly. Then, we need to cast to char to avoid compiler's warning of a 'possible loss of data'. Since that we are storing values which are not greater than 256, we are sure that no data is lost.
Unfortunately, the retrieved value, as shown by debugger, is not 130, but -126.
If you think that this is the bug I'm talking about, it is not. This is just a usual signed/unsigned issue. Since c is declared as a (signed) char, the moment the 1 hits its upper bit it becomes a negative value. This is not a bug - this is how it supposed to be. The bits are all the same. It is still 130 under the hood. Let's continue.
Retrieve the same value as an integer:
int i = (short)f->Item["MyField"]->Value;
As you probably guessed, this time the debugger shows 130.
Now, what do you think of this code:
if (c == i)
{
AfxMessageBox("Heaven");
}
else
{
AfxMessageBox("Hell");
}
Which message will we get? On the one hand, this is a case of -126 vs. 130. On the other, we know that:
- Both values have same bits - 10000010, which are read from the same database field.
- Both values are signed values, so there is no signed/unsigned mismatch here.
- When evaluating the expression, both values are converted to same type, int. So when compared, they are of same size. In other words, compiler sees if ((int)c == i)<code>.
Which message will you see? I have to tell you. Despite of these assumptions, you'll be getting the else branch. This is what we call Potekhin's bug.
Explanation of the trick
One may say, of course, 130 is not -126, and that's why they don't match. However, this does not explain how did we end up with such results. Here's the explanation. Compare:
Before conversion to int:
(int) 130 == 10000010 (char) -126 == 10000010
After conversion to int:
(int) 130 == 10000010 (int) -126 == 11111111111111111111111110000010
When a char gets converted, its negative bit gets propagated all the way to the left.
Well, I knew it. Kind of. I knew that it gets propagated. The problem is, I didn't realize that it could lead to scenarios like the described above.
Conclusion? Same old rule. Never use a signed char to store anything above 128. Or if you do, don't compare it to an int :)
