• Jesse Phillips (10/10) Nov 22 2011 There is a good possibility that I don't know anything, but is there som...
• Jerry (13/23) Nov 25 2011 In this case, T1 (a) is a signed value and b is unsigned. To do the

Jesse Phillips <Jesse.K.Phillips gmail.com> writes:

There is a good possibility that I don't know anything, but is there something
about doing two checks that goes faster?

        static if (isIntegral!(T1) && isIntegral!(T2)
                   && (mostNegative!(T1) < 0) != (mostNegative!(T2) < 0))
            static if (mostNegative!(T1) < 0)
                immutable chooseB = b > a || a < 0;
            else
                immutable chooseB = b > a && b > 0;
        else
            immutable chooseB = b > a;

What has me baffled is the second checks after b > a. Doesn't b > a always
answer the question? The only thing I can think of is a check for negative
being faster, so shouldn't they come before the b > a check?

Jerry <jlquinn optonline.net> writes:

Jesse Phillips <Jesse.K.Phillips gmail.com> writes:

 There is a good possibility that I don't know anything, but is there something
about doing two checks that goes faster?

         static if (isIntegral!(T1) && isIntegral!(T2)
                    && (mostNegative!(T1) < 0) != (mostNegative!(T2) < 0))
             static if (mostNegative!(T1) < 0)
                 immutable chooseB = b > a || a < 0;


In this case, T1 (a) is a signed value and b is unsigned.  To do the
comparison, D will implicitly convert the signed variable 'a' to
unsigned.  This will wrap around the negative values.  Therefore, using the
case of 32 bit ints as an example, if b = 1, and a = -1, the comparison
performed will be

          1 > 0xffffffff || -1 < 0

The first test fails, so the 2nd is required to succeed.

             else
                 immutable chooseB = b > a && b > 0;

In this case, a is the unsigned value and b is signed and cast to
unsigned byt the compiler.  With a = 1 and b = -1, the test becomes

         0xffffffff > 1 && -1 > 0

The 2nd test is required to make this fail.

         else
             immutable chooseB = b > a;

 What has me baffled is the second checks after b > a. Doesn't b > a always
answer the question? The only thing I can think of is a check for negative
being faster, so shouldn't they come before the b > a check?


Jerry Quinn