Re: How can I get rid of "Warning C4307: integral constant overflow"
On Jun 5, 10:49 pm, "Victor Bazarov" <v.Abaza...@comAcast.net> wrote:
_Christopher(M2M) wrote:
Microsoft Visual Studio 2005 Version 8.0.50727.762 (SP.050727-7600)
How do I get rid of "Warning C4307: '-' : integral constant overflow"
from the following macro?
#define d_ReturnMax(I_iTypeSize)\
( (int64(1)<<((I_iTypeSize*8)-1))-1)
There NO warnings when done as a function:
int64 d_ReturnMax(int64 I_iTypeSize)
{ return (1<<((I_iTypeSize*8)-1))-1;}
Test code:
int64 L_iMaxIntSize=d_ReturnMax(sizeof(int32)); //
L_iMaxIntSize=0x000000007fffffff
int64 L_iMaxIntSize=d_ReturnMax(sizeof(char)); //
L_iMaxIntSize=0x000000000000007f
I tried typecasting around all parts of code i.e.
int64(...)/(int64)... I give up now.
Can anyone else spot something I have missed?
Probably. One, 'int64' is compiler-specific and should be asked
about in the newsgroup dedicated to that compiler.
int64_t is part of C99, and is in the current draft. It's
pretty obvious, I think, what int64 is supposed to be.
Two, to form
a constant of a particular type you need to follow it with a certain
suffix, like 0xffL for 'long' or 0x55U for 'unsigned' (hint: int64
probably has its own dedicated suffix), do not use a "cast".
There's no problem using a cast as long as the value is in fact
representable as an int. In his case, the value is 1, so
there's no problem. In many cases, the cast is necessary,
because you don't know the actual type involved, e.g. if dealing
with size_t.
Also,
shifts only work with the right argument between 0 and some
(relatively small) number.
Make sure whatever '((I_iTypeSize*8)-1))'
(do you really need all those parentheses?) expands into is between
0 and the maximum allowed value (63?).
In his test code, the conditions are met. Or at least they are
if int64 really is a 64 bit int; the only explination I can see
is that the compiler is evaluating the integral constant
expression as a 32 bit long, despite the presence of an int64 in
it. (Perhaps this part of the compiler hasn't been reworked to
reflect the presence of the new integral types.) At any rate,
there is no overflow in his expression.
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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