Hi
if have simple macro defined this way:
#define M_OPplus( classname ) typedef classname operator+(const
classname& rhs);
and a template class :
template<typena me X, typename Y>
class Test1
{
public:
double run();
M_OPplus( Test1<X,Y)
};
The compiler (VS2003) is complaining : warning C4002: too many actual
parameters for macro, . I think it is related to the comma in the
middle , the preprocessor may think that i am passing two parameters
to the macro.
Is it possible to fix it and does it conform with the C++ standart?
Many Thx
12  3229  sw*****@googlem ail.com wrote:
Hi
if have simple macro defined this way:
#define M_OPplus( classname ) typedef classname operator+(const
classname& rhs);
and a template class :
template<typena me X, typename Y>
class Test1
{
public:
double run();
M_OPplus( Test1<X,Y)
^
the comma split up your argument into to parts
write
M_OPplus(( Test1<X,Y))
};
The compiler (VS2003) is complaining : warning C4002: too many actual
parameters for macro, . I think it is related to the comma in the
middle , the preprocessor may think that i am passing two parameters
to the macro.
Is it possible to fix it and does it conform with the C++ standart?
Many Thx
--
Thanks
Barry
<sw*****@google mail.comwrote in message
news:11******** **************@ y42g2000hsy.goo glegroups.com.. .
if have simple macro defined this way:
#define M_OPplus( classname ) typedef classname operator+(const
classname& rhs);
and a template class :
template<typena me X, typename Y>
class Test1
{
public:
double run();
M_OPplus( Test1<X,Y)
};
The compiler (VS2003) is complaining : warning C4002: too many actual
parameters for macro, . I think it is related to the comma in the
middle , the preprocessor may think that i am passing two parameters
to the macro.
Your correct.
Is it possible to fix it and does it conform with the C++ standart?
Yes. You can use a little trick that involves passing a name of a macro
function to the M_OPplus macro which calls it to get at the tokens.
Something like this:
#define M_OPplus(classn ame_macro) \
typedef classname_macro () \
operator+(const classname_macro ()& rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
See what I am getting at here?
"Barry" <dh*****@gmail. comwrote in message news:fc******** **@aioe.org...
sw*****@googlem ail.com wrote:
[...]
the comma split up your argument into to parts
write
M_OPplus(( Test1<X,Y))
[...]
That introduces extra parenthesis which can muck up syntax. For instance:
This has a syntax error:
_______________
#define M_OPplus(classn ame) \
typedef classname \
operator+(const classname &rhs)
template<typena me X, typename Y>
class Test1 {
public:
double run();
M_OPplus((Test1 <X,Y>));
};
_______________
This does not:
_______________
#define M_OPplus(classn ame_macro) \
typedef classname_macro () \
operator+(const classname_macro ()& rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
_______________
Chris Thomasson wrote:
"Barry" <dh*****@gmail. comwrote in message news:fc******** **@aioe.org...
>sw*****@googlem ail.com wrote:
[...]
>the comma split up your argument into to parts
write
M_OPplus(( Test1<X,Y))
[...]
That introduces extra parenthesis which can muck up syntax. For instance:
This has a syntax error:
_______________
#define M_OPplus(classn ame) \
typedef classname \
operator+(const classname &rhs)
template<typena me X, typename Y>
class Test1 {
public:
double run();
M_OPplus((Test1 <X,Y>));
};
_______________
This does not:
_______________
#define M_OPplus(classn ame_macro) \
typedef classname_macro () \
operator+(const classname_macro ()& rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
_______________
yeh, you are right,
didn't carefully view the context
--
Thanks
Barry
Chris Thomasson wrote:
<sw*****@google mail.comwrote in message
news:11******** **************@ y42g2000hsy.goo glegroups.com.. .
>if have simple macro defined this way:
#define M_OPplus( classname ) typedef classname operator+(const
classname& rhs);
and a template class :
template<typen ame X, typename Y>
class Test1
{
public:
double run();
M_OPplus( Test1<X,Y)
};
The compiler (VS2003) is complaining : warning C4002: too many actual
parameters for macro, . I think it is related to the comma in the
middle , the preprocessor may think that i am passing two parameters
to the macro.
Your correct.
>Is it possible to fix it and does it conform with the C++ standart?
Yes. You can use a little trick that involves passing a name of a macro
function to the M_OPplus macro which calls it to get at the tokens.
Something like this:
#define M_OPplus(classn ame_macro) \
typedef classname_macro () \
operator+(const classname_macro ()& rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
#define M_Test1 ....
definition and invocation should be identical
public:
double run();
M_OPplus(M_Test 1);
};
See what I am getting at here?
--
Thanks
Barry
"Barry" <dh*****@gmail. comwrote in message news:fc******** **@aioe.org...
Chris Thomasson wrote:
[...]
>#define M_OPplus(classn ame_macro) \
typedef classname_macro () \
operator+(const classname_macro ()& rhs)
template<typen ame X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
#define M_Test1 ....
definition and invocation should be identical
[...]
I am not exactly sure what you mean here. Are you suggesting that 'M_Test1'
does not really need to be a macro function? Well, in that case the code
would need to be modified to something like:
____________
#define M_OPplus(classn ame) \
typedef classname \
operator+(const classname &rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1 Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
____________
At first glance I thought that the 'M_Test1' macro would be expanded and
then passed to 'M_OPplus' which would produce the original error that the OP
was asking about. After I compiled it (gcc) I realized that it was getting
expanded in the context of the 'M_OPplus'.
I guess the only drawback from this would be that you could not pass the
'classname' parameter to another similar macro function. For instance:
this does not compile:
____________
#define M_Typedef(class name) \
typedef classname
#define M_OPplus(classn ame) \
M_Typedef(class name) \
operator+(const classname &rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1 Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
____________
while this does:
____________
#define M_Typedef(class name_macro) \
typedef classname_macro ()
#define M_OPplus(classn ame_macro) \
M_Typedef(class name_macro) \
operator+(const classname_macro ()& rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
____________
So I guess defining 'M_Test1' as a macro function instead of a plain macro
is more "flexible".
Any thoughts?
Chris Thomasson wrote:
"Barry" <dh*****@gmail. comwrote in message news:fc******** **@aioe.org...
>Chris Thomasson wrote:
[...]
>>#define M_OPplus(classn ame_macro) \
typedef classname_macro () \
operator+(const classname_macro ()& rhs)
template<type name X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
#define M_Test1 ....
definition and invocation should be identical
[...]
I am not exactly sure what you mean here. Are you suggesting that
'M_Test1' does not really need to be a macro function? Well, in that
case the code would need to be modified to something like:
I mean if you
#define M_Test1() ...
then call
M_Test1() other than M_Test1
if you
#define M_Test1 ...
then call
M_Test1 other than M_Test1()
--
Thanks
Barry
"Barry" <dh*****@gmail. comwrote in message news:fc******** **@aioe.org...
Chris Thomasson wrote:
>"Barry" <dh*****@gmail. comwrote in message
news:fc******* ***@aioe.org...
>>Chris Thomasson wrote:
[...]
>I am not exactly sure what you mean here. Are you suggesting that
'M_Test1' does not really need to be a macro function? Well, in that case
the code would need to be modified to something like:
[...]
I mean if you
#define M_Test1() ...
then call
M_Test1() other than M_Test1
Well, as soon as I call M_Test1() it will expand on the spot. I want to be
able to delay expansion until the exact place I need it. Therefore, I can
use the _name_ of the macro function M_Test1 as a sort-of function pointer.
This is the "trick" I mentioned to the OP...
Here is an example program you can compile:
_______________
#include <cstdio>
#define CALL_MACRO_FUNC TION(func_ptr)f unc_ptr()
#define MY_MESSAGE() "Press <ENTERto exit."
int main(void) {
puts(CALL_MACRO _FUNCTION(MY_ME SSAGE));
getchar();
return 0;
}
_______________
As far as I can tell, this is 100% legitimate, and conforms to the standard.
Does that make sense to you?
Chris Thomasson wrote:
"Barry" <dh*****@gmail. comwrote in message news:fc******** **@aioe.org...
>Chris Thomasson wrote:
[...]
>>#define M_OPplus(classn ame_macro) \
typedef classname_macro () \
operator+(const classname_macro ()& rhs)
template<type name X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
#define M_Test1 ....
definition and invocation should be identical
[...]
I am not exactly sure what you mean here. Are you suggesting that
'M_Test1' does not really need to be a macro function? Well, in that
case the code would need to be modified to something like:
____________
#define M_OPplus(classn ame) \
typedef classname \
operator+(const classname &rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1 Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
____________
At first glance I thought that the 'M_Test1' macro would be expanded and
then passed to 'M_OPplus' which would produce the original error that
the OP was asking about. After I compiled it (gcc) I realized that it
was getting expanded in the context of the 'M_OPplus'.
I guess the only drawback from this would be that you could not pass the
'classname' parameter to another similar macro function. For instance:
this does not compile:
____________
#define M_Typedef(class name) \
typedef classname
#define M_OPplus(classn ame) \
M_Typedef(class name) \
operator+(const classname &rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1 Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
____________
while this does:
____________
#define M_Typedef(class name_macro) \
typedef classname_macro ()
#define M_OPplus(classn ame_macro) \
M_Typedef(class name_macro) \
operator+(const classname_macro ()& rhs)
template<typena me X, typename Y>
class Test1 {
#define M_Test1() Test1<X,Y>
public:
double run();
M_OPplus(M_Test 1);
};
____________
So I guess defining 'M_Test1' as a macro function instead of a plain
macro is more "flexible".
Well, forgive my English
You know
#define SOME_MACRO()
int main()
{
SOME_MACRO();
SOME_MACRO; // does not compile
}
so IMHO,
the definition and invoking of macro should better give the same form.
--
Thanks
Barry This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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