A strange operation problem in SQL Server: -100/-100*10 = 0

According to the precedence table, this is the expected behavior. The operator with higher precedence (/ and *) is evaluated before operator with lower precedence (unary -). So this:

-100 / -100 * 10

is evaluated as:

-(100 / -(100 * 10))

Note that this behavior is different from most programming languages where unary negation has higher precedence than multiplication and division e.g. VB, JavaScript.

BOL is correct. - has lower precedence than *, so

-A * B

is parsed as

-(A * B)

Multiplication being what it is, you don't typically notice this, except when mixing in the two other binary operators with equal precedence: / and % (and % is rarely used in compound expressions like this). So

C / -A * B

Is parsed as

C / -(A * B)

explaining the results. This is counter-intuitive because in most other languages, unary minus has higher precedence than * and /, but not in T-SQL, and this is documented correctly.

A nice (?) way to illustrate it:

SELECT -1073741824 * 2

produces an arithmetic overflow, because -(1073741824 * 2) produces 2147483648 as an intermediate, which does not fit in an INT, but

SELECT (-1073741824) * 2

produces the expected result -2147483648, which does.

Notice in the documentation that (perhaps counter-intuitively) the order of precedence for - (Negative) is third.

So you effectively get:

-(100/-(100*10)) = 0

If you place them into variables you won't see this happening, as there is no unary operation that occurs after the multiplication.

So here A and B are the same, whereas C, D, E show the result you are seeing (with E having the complete bracketing)

DECLARE @i1 int, @i2 int, @i3 int;

SELECT @i1 = -100,
       @i2 = -100,
       @i3 = 10;

SELECT @i1/@i2*@i3      [A],
       -100/(-100)*10   [B],
       -100/-100*10     [C],
       -100/-(100*10)   [D],
       -(100/-(100*10)) [E];

A - 10
B - 10
C - 0
D - 0
E - 0