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cody
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 Posted: Sun Jul 13, 2003 6:01 am Post subject: Re: Official c++ 64-bit integer data type |
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| Quote: | Unfortunatly using the same names will not increase portability where
it matters:
int32_t = int16_t + int16_t;
This example will have different results depending on the sizeof(int).
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you add datatypes together????
what do you mean here?
--
cody
[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
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Michiel Salters
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| Posted: Mon Jul 14, 2003 7:10 pm Post subject: Re: Official c++ 64-bit integer data type |
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[email]deutronium (AT) gmx (DOT) net[/email] ("cody") wrote in message news:<beqfj3$7u1ck$1 (AT) ID-176797 (DOT) news.uni-berlin.de>...
| Quote: | Unfortunatly using the same names will not increase portability where
it matters:
int32_t = int16_t + int16_t;
This example will have different results depending on the sizeof(int).
you add datatypes together????
what do you mean here?
|
Adding variables of the specified type together, e.g.
int16_t a = 1<<14;
int16_t b = 1<<14;
int32_t c = a + b;
The obvious result would be c = 1<<15, but in the addition
the types will be promoted to int. The result of the addition
will also be an int. Whether or not that overflows still
depends on the exact bitwidth of an int. c==-(1<<15) could
occur if ints are 16 bits.
Regards,
--
Michiel Salters
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Richard Corden
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| Posted: Mon Jul 14, 2003 9:58 pm Post subject: Re: Official c++ 64-bit integer data type |
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| Quote: | int32_t = int16_t + int16_t;
This example will have different results depending on the sizeof(int).
you add datatypes together????
what do you mean here?
|
Sorry for the poor example...
void foo (int16_t a, int16_t b)
{
int32_t result;
result = a + b;
}
So, the value of result may be different for different sizes of int.
Regards
Richard.
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Richard Corden
To reply remove 's' from address
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cody
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| Posted: Thu Jul 17, 2003 1:43 am Post subject: Re: Official c++ 64-bit integer data type |
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| Quote: | void foo (int16_t a, int16_t b)
{
int32_t result;
result = a + b;
}
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you mean in the case if int would be 8 bits then:
300 + 200 = 88 and not 600
because every integral expression where not type ios explicitely specified
results int the int-type?
--
cody
[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
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Philippe Mori
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| Posted: Mon Jul 21, 2003 6:32 pm Post subject: Re: Official c++ 64-bit integer data type |
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""cody"" <deutronium (AT) gmx (DOT) net> a écrit dans le message de
news:bf466n$auf1n$1 (AT) ID-176797 (DOT) news.uni-berlin.de...
| Quote: | void foo (int16_t a, int16_t b)
{
int32_t result;
result = a + b;
}
you mean in the case if int would be 8 bits then:
300 + 200 = 88 and not 600
because every integral expression where not type ios explicitely specified
results int the int-type?
--
|
If you want portable code then you would write something like:
int32_t foo(int16_t a, int16_t bn)
{
int32_t result = a;
result += b;
return result;
}
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cody
Guest
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| Posted: Tue Jul 22, 2003 10:56 pm Post subject: Re: Official c++ 64-bit integer data type |
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| Quote: | If you want portable code then you would write something like:
int32_t foo(int16_t a, int16_t bn)
{
int32_t result = a;
result += b;
return result;
}
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better would be:
inline int32_t foo(int16_t a, int16_t bn)
{
return (int32_t)a + b;
}
:-D
--
cody
[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
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