I'm writing a template, and I need to know the maximum value of a given
integer type at compile time. something like:
template<class NumType>
class Arb {
public:
static NumType const max = /* maximum value */;
};
If the template were only to be used with unsigned integer types, then
I'd do the following:
template<class NumType>
class Arb {
public:
static NumType const max = -1;
};
I need a compile-time constant which evaluates to the maximum value of an
integer type.
Any ideas?
If it weren't undefined behaviour to overflow a signed integer, I could
use a metaprogramming technique such as the following:
template<class T, bool overflow = false>
struct MaxIntVal {
private:
static T const internal = 1 + MaxIntVal<T,...
public:
static T const val = SCHAR_MAX + internal;
};
--
Frederick Gotham
Jul 4 '06
35  12588 
* Frederick Gotham:
Alf P. Steinbach posted:
>Note the bit pattern for the highest value when using a given number of
bits.
I was under the assumption that C++ had to store positive values as
follows:
0: 0000
1: 0001
2: 0010
3: 0011
4: 0100
5: 0101
6: 0110
7: 0111
8: 1000
9: 1001
10: 1010
11: 1011
12: 1100
13: 1101
14: 1110
15: 1111
Are you saying that:
(A) The machine could use another method such as "Gray code".
and
(B) My code could break on other systems, and thus isn't trully
portable?
It's just academic, but yes. §5.8/2 defines the result of << as a
bit-level operation, and as a value only for unsigned types. §5.8/3
defines the result of >as a bit-level operation and as a value both
for signed and unsigned types (the latter only for non-negative values).
The cases excluded from these definitions are, AFAICT, exactly those
where the definitions would otherwise be incompatible with grey-code...
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
* Rolf Magnus:
Rolf Magnus wrote:
>Alf P. Steinbach wrote:
>>It's good fun to throw academic spanners into practical wheels, so,
where does the standard exclude a grey-code representation? ;-)
Hmm, I would have said 3.9.1/7, but after reading it again, it seems that
this actually doesn't prohibit a grey-code representation.
Ok, now I have one: 5.8/2.
Nope, not as far as I can see: it's /very/ ingeniously worded so that
the cases that could be incompatible are simply not defined.
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
* Alf P. Steinbach:
grey-code
Sorry, again: _gray code_.
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Frederick Gotham wrote:
Rolf Magnus posted:
>Ok, now I have one: 5.8/2.
The value of E1 << E2 is E1 (interpreted as a bit pattern) left-shifted E2
bit positions; vacated bits are zero-filled. If E1 has an unsigned type,
the value of the result is E1 multiplied by the quantity 2 raised to the
power E2, reduced modulo ULONG_MAX+1 if E1 has type unsigned long,
UINT_MAX+1 otherwise.
Is it the "vacated bits are zero-filled" part that you're talking about?
No, it's the "(interpret ed as a bit pattern) left-shifted E2 bit positions"
in combination with the "multiplied by the quantity 2 raised to the power
E2". If your values are represented in gray code, a left-shift of the bit
pattern doesn't necessarily result in a mulitplication by 2.
* Rolf Magnus:
Frederick Gotham wrote:
>Rolf Magnus posted:
>>Ok, now I have one: 5.8/2.
The value of E1 << E2 is E1 (interpreted as a bit pattern) left-shifted E2
bit positions; vacated bits are zero-filled. If E1 has an unsigned type,
the value of the result is E1 multiplied by the quantity 2 raised to the
power E2, reduced modulo ULONG_MAX+1 if E1 has type unsigned long,
UINT_MAX+1 otherwise.
Is it the "vacated bits are zero-filled" part that you're talking about?
No, it's the "(interpret ed as a bit pattern) left-shifted E2 bit positions"
in combination with the "multiplied by the quantity 2 raised to the power
E2". If your values are represented in gray code, a left-shift of the bit
pattern doesn't necessarily result in a mulitplication by 2.
That's only for unsigned types. C++ unsigned integers are pure binary.
For signed integers there's much more leeway (formally).
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Alf P. Steinbach wrote:
* Rolf Magnus:
>Rolf Magnus wrote:
>>Alf P. Steinbach wrote:
It's good fun to throw academic spanners into practical wheels, so,
where does the standard exclude a grey-code representation? ;-)
Hmm, I would have said 3.9.1/7, but after reading it again, it seems
that this actually doesn't prohibit a grey-code representation.
Ok, now I have one: 5.8/2.
Nope, not as far as I can see: it's /very/ ingeniously worded so that
the cases that could be incompatible are simply not defined.
What's not defined? It says clearly that the bit pattern is shifted left by
the specified number of bits and that this will (for unsigned types) result
in a multiplication by 2 raised to the value of the rhs. That's not
possible with Gray code, so unsigned types cannot use Gray code. Since a
bit pattern for a signed value must be the same as for the corresponding
unsigned value, signed types neither can use Gray code.
Alf P. Steinbach wrote:
* Rolf Magnus:
>Frederick Gotham wrote:
>>Rolf Magnus posted:
Ok, now I have one: 5.8/2.
The value of E1 << E2 is E1 (interpreted as a bit pattern) left-shifted
E2 bit positions; vacated bits are zero-filled. If E1 has an unsigned
type, the value of the result is E1 multiplied by the quantity 2 raised
to the power E2, reduced modulo ULONG_MAX+1 if E1 has type unsigned
long, UINT_MAX+1 otherwise.
Is it the "vacated bits are zero-filled" part that you're talking about?
No, it's the "(interpret ed as a bit pattern) left-shifted E2 bit
positions" in combination with the "multiplied by the quantity 2 raised
to the power E2". If your values are represented in gray code, a
left-shift of the bit pattern doesn't necessarily result in a
mulitplicati on by 2.
That's only for unsigned types. C++ unsigned integers are pure binary.
For signed integers there's much more leeway (formally).
Only for the negative values. So I guess a signed integer could use Gray
code for the negative part. That would be extremely odd, but seems to be
allowed by the standard.
* Rolf Magnus:
Alf P. Steinbach wrote:
>* Rolf Magnus:
>>Frederick Gotham wrote:
Rolf Magnus posted:
Ok, now I have one: 5.8/2.
The value of E1 << E2 is E1 (interpreted as a bit pattern) left-shifted
E2 bit positions; vacated bits are zero-filled. If E1 has an unsigned
type, the value of the result is E1 multiplied by the quantity 2 raised
to the power E2, reduced modulo ULONG_MAX+1 if E1 has type unsigned
long, UINT_MAX+1 otherwise.
Is it the "vacated bits are zero-filled" part that you're talking about?
No, it's the "(interpret ed as a bit pattern) left-shifted E2 bit
positions" in combination with the "multiplied by the quantity 2 raised
to the power E2". If your values are represented in gray code, a
left-shift of the bit pattern doesn't necessarily result in a
mulitplicatio n by 2.
That's only for unsigned types. C++ unsigned integers are pure binary.
For signed integers there's much more leeway (formally).
Only for the negative values. So I guess a signed integer could use Gray
code for the negative part. That would be extremely odd, but seems to be
allowed by the standard.
No, it's for any (signed) value. Left shift doesn't define the numeric
value. And right shift >does define the value but is all OK.
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
* Rolf Magnus:
Alf P. Steinbach wrote:
>* Rolf Magnus:
>>Rolf Magnus wrote:
Alf P. Steinbach wrote:
It's good fun to throw academic spanners into practical wheels, so,
where does the standard exclude a grey-code representation? ;-)
Hmm, I would have said 3.9.1/7, but after reading it again, it seems
that this actually doesn't prohibit a grey-code representation.
Ok, now I have one: 5.8/2.
Nope, not as far as I can see: it's /very/ ingeniously worded so that
the cases that could be incompatible are simply not defined.
What's not defined? It says clearly that the bit pattern is shifted left by
the specified number of bits and that this will (for unsigned types) result
in a multiplication by 2 raised to the value of the rhs. That's not
possible with Gray code, so unsigned types cannot use Gray code. Since a
bit pattern for a signed value must be the same as for the corresponding
unsigned value, signed types neither can use Gray code.
No, the bit pattern for a signed value must not necessarily be the same
as for the corresponding unsigned value. In particular, that very
obviously doesn't hold for negative values. And by extension, it
doesn't hold for non-negative values either (if you're looking at the
same paragraph which has been discussed to death in this connection
before [the solution, if it may be called that, is then to look at the
definitions of terms] ;-)).
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Alf P. Steinbach wrote:
* Rolf Magnus:
>Alf P. Steinbach wrote:
>>* Rolf Magnus:
Rolf Magnus wrote:
Alf P. Steinbach wrote:
>
>It's good fun to throw academic spanners into practical wheels, so,
>where does the standard exclude a grey-code representation? ;-)
Hmm, I would have said 3.9.1/7, but after reading it again, it seems
that this actually doesn't prohibit a grey-code representation.
Ok, now I have one: 5.8/2.
Nope, not as far as I can see: it's /very/ ingeniously worded so that
the cases that could be incompatible are simply not defined.
What's not defined? It says clearly that the bit pattern is shifted left
by the specified number of bits and that this will (for unsigned types)
result in a multiplication by 2 raised to the value of the rhs. That's
not possible with Gray code, so unsigned types cannot use Gray code.
Since a bit pattern for a signed value must be the same as for the
correspondin g unsigned value, signed types neither can use Gray code.
No, the bit pattern for a signed value must not necessarily be the same
as for the corresponding unsigned value. In particular, that very
obviously doesn't hold for negative values. And by extension, it
doesn't hold for non-negative values either (if you're looking at the
same paragraph which has been discussed to death in this connection
before [the solution, if it may be called that, is then to look at the
definitions of terms] ;-)).
You must be talking about :
"The range of nonnegative values of a signed integer type is a subrange of
the corresponding unsigned integer type, and the value representation of
each corresponding signed/unsigned type shall be the same."
Not sure what's wrong about that, though. Doesn't it mean that the same
value in a corresponding signed and unsigned type must have the same bit
pattern? Obviously, that's only possible for the subrange that is covered
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