Slightly surprised that the following didn't compile
#inclue <algorithm>
enum { C = 10 };
int main()
{
char a[C];
std::fill_n(a, C, '\0');
}
Error message points to implementation of std::fill_n and complains that
unary '--' : '' does not define this operator or a conversion to a type
acceptable to the predefined operator
Essentially I think its complaining that it can't modify the value of the
enum C.
Is this correct? I expected the template to be instantiated with an int
which could be modified using operator--.
Replacing enum { C = 10 }; with const int C = 10; does compile.
John 12 6098
On Tue, 15 Jul 2003, John Harrison wrote: Slightly surprised that the following didn't compile
#inclue <algorithm> enum { C = 10 }; int main() { char a[C]; std::fill_n(a, C, '\0'); }
Error message points to implementation of std::fill_n and complains that
unary '--' : '' does not define this operator or a conversion to a type acceptable to the predefined operator
Essentially I think its complaining that it can't modify the value of the enum C.
Is this correct? I expected the template to be instantiated with an int which could be modified using operator--.
Replacing enum { C = 10 }; with const int C = 10; does compile.
John
Your code, as above, compiles, as is, with gcc 3.2.2. Did you use fill()
instead of fill_n()?
"Jesse Nowells" <jn******@trans bay.net> wrote in message
news:Pine.BSF.4 .31.03071500531 30.80673-100000@localhos t...
On Tue, 15 Jul 2003, John Harrison wrote:
Slightly surprised that the following didn't compile
#inclue <algorithm> enum { C = 10 }; int main() { char a[C]; std::fill_n(a, C, '\0'); }
Error message points to implementation of std::fill_n and complains that
unary '--' : '' does not define this operator or a conversion to a type acceptable to the predefined operator
Essentially I think its complaining that it can't modify the value of
the enum C.
Is this correct? I expected the template to be instantiated with an int which could be modified using operator--.
Replacing enum { C = 10 }; with const int C = 10; does compile.
John
Your code, as above, compiles, as is, with gcc 3.2.2. Did you use fill() instead of fill_n()?
No I used code exactly as above, cut and paste from my compiler, except for
#include <algorithm> which I managed to mistype.
Compiler is VC++ .NET
john
"Alf P. Steinbach" <al***@start.no > wrote in message
news:3f******** ********@News.C IS.DFN.DE... On Tue, 15 Jul 2003 08:03:01 +0100, "John Harrison"
<jo************ *@hotmail.com> wrote:Slightly surprised that the following didn't compile
#inclue <algorithm> Hah, there's your culprit right there. ;-)
The rest of the code was cut and paste, I swear!
enum { C = 10 }; int main() { char a[C]; std::fill_n(a, C, '\0'); }
Error message points to implementation of std::fill_n and complains that
unary '--' : '' does not define this operator or a conversion to a type acceptable to the predefined operator
enum E{ C = 10 };
E operator--( E x ){ return static_cast<E>( x-1 ); } Essentially I think its complaining that it can't modify the value of the enum C.
Nope. It complains that enum type doesn't have a decrement operator.
So presumably this would also compile (don't have my compiler handy to
check)
enum { C = 10 };
int main()
{
char a[C];
std::fill_n(a, (int)C, '\0');
}
john
On Tue, 15 Jul 2003 09:34:34 +0100, "John Harrison" <jo************ *@hotmail.com> wrote: So presumably this would also compile (don't have my compiler handy to check)
enum { C = 10 }; int main() { char a[C]; std::fill_n(a, (int)C, '\0'); }
It does.
What happens if the give the enum type a name? This made it fail to compile
with BCB4.
Fraser.
John Harrison wrote: Essentially I think its complaining that it can't modify the value of the enum C.
Is this correct? I expected the template to be instantiated with an int which could be modified using operator--.
There is no rule excluding enum types from being the first class
citizens when it comes to instantiation of templates.
There is another issue with calling -- on enum types. According to the
C++ standard (5.3.2) one can call the (built-in) prefix operator ++ or
-- on a operand being an lvalue of an arithmetic or pointer to
completely-defined object type. No enum is mentioned in this context
however it is when i.e., the unary +/- operators are discussed.
Nevertheless many compilers will accept such an operation mainly because
of their own backward compatibility so it may be really hard to spot.
Regards,
Janusz
On Tue, 15 Jul 2003 09:33:50 -0400, "Ron Natalie" <ro*@sensor.com > wrote: "Alf P. Steinbach" <al***@start.no > wrote in message news:3f******** ********@News.C IS.DFN.DE...
E operator--( E x ){ return static_cast<E>( x-1 ); }
-x is not x-1
E is not necessarily able to hold the value -10. Nope. It complains that enum type doesn't have a decrement operator.
No it complains about UNARY -.
Hello? Anyone home today, Ron? Must be the heat.
John Harrison wrote: That's very strange because since enums are often used for ordinal types you would think that if any arithmetic operators were to be defined then it would be ++ and --. Just an oversight I guess.
Nope, deliberate. We didn't want to require compiler writers to generate
increment and decrement code for things like this:
enum flags { 0x01, 0x02, 0x04, 0x08, 0x10 };
--
"To delight in war is a merit in the soldier,
a dangerous quality in the captain, and a
positive crime in the statesman."
George Santayana
"Bring them on."
George W. Bush
John Harrison wrote: "Pete Becker" <pe********@acm .org> wrote in message news:3F******** *******@acm.org ... John Harrison wrote: That's very strange because since enums are often used for ordinal types you would think that if any arithmetic operators were to be defined then it would be ++ and --. Just an oversight I guess.
Nope, deliberate. We didn't want to require compiler writers to generate increment and decrement code for things like this:
enum flags { 0x01, 0x02, 0x04, 0x08, 0x10 };
OK I can buy that but then why define unary minus?
--, ++, +=, etc. all must create values of the same type as their
operand. Unary minus, as well as the rest of the arithmetic operators,
create integral values. Read about the "usual arithmetic conversions."
--
"To delight in war is a merit in the soldier,
a dangerous quality in the captain, and a
positive crime in the statesman."
George Santayana
"Bring them on."
George W. Bush This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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