Big hex numbers negative?

Fabian Cenedese Cenedese at
Tue Feb 24 01:13:25 CST 2004

>> >double dVal;
>> >OLECHAR test1[]={'&', 'H', '8', '0', '0', '0', '0', '0', '0', '0', '\0'};
>> >ok=VarR8FromStr(test1, LANG_NEUTRAL, NUMPRS_STD, &dVal);
>> >
>> >The result for dVal was -2147483648. But as a real value it shouldn't
>> >have any problems holding the "real" value 2147483648. So why has
>> >it become negative? Is it because the source form was a hex number?
>> >Are all hex numbers automatically signed if converted to int/real? Or
>> >just because of the 32nd bit? The documentation wasn't that informative.
>> >
>> >(The funny thing though was this remark in my VC6 help, it's not
>> >in the online version of MSDN anymore:
>> >Passing into this function any invalid and, under some circumstances, NULL
>> > pointers will result in unexpected termination of the application. For
>> > more information about handling exceptions, see Programming
>> > Considerations.
>> >
>Now does that mean "invalid numbers" or "invalid pointers"?  Are they just 
>saying that they hadn't included IsBadReadPtr in the argument checking?

I don't know, that was the only remark. But the same remark is included
in all VarXXFromYY functions.

>> I'm not sure I understood your question properly.
>> A signed 2 complement 32 bit var can hold the numbers (-2^31) to
>> (2^31)-1. That's just how the encoding works. Was that your question?
>>              Shachar
>Because there is nothing explicitly in there about 32 bit representations.  It 
>goes from a string of characters to a (excuse my ignorance) 64 bit double 
>prescision.  So are there restrictions on what character strings can be 
>passed in?  e.g. what would it do with "&H80000000"?

My original question was if this conversion works implicitly with a signed
hex string even if hex can be anything. Strings with more than 8 significant
chars are rejected (DISP_E_OVERFLOW). So I guess these conversions
assume at the most a signed long. That's how I will implement them then.


bye  Fabi

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