I am baffled about how to use mixins.  In the online documentation for mixin's there is the following statement:

" Unlike a template instantiation, a template mixin's body is evaluated within the scope where the mixin appears, not where the template declaration is defined. It is analogous to cutting and pasting the body of the template into the location of the mixin. It is useful for injecting parameterized 'boilerplate' code, as well as for creating templated nested functions, which is not possible with template instantiations."

Unfortunately, the current version of the compiler does not evalute the mixin where it is declared per se,  in that all declarations local to the location of the mixin, but not visible to the template declaration cause compile errors. For example consider the following code, which attempts to select one of two possible computations:

template Foo(int select: 1) {
x = y + z - s.b ;
}
template Foo(int select: 0) {
x = y + z + s.b ;
}

struct stuff {
double b = 5 ;
}

int main(char[][] args)
{
stuff s ;
double x, y = 2.0, z = 3.0 ;
mixin Foo!(1) ;
printf("x = %g\n", x) ;
return(0) ;
}

This fails to compile with the error
mixintest.d(2): no identifier for declarator x
mixintest.d(5): no identifier for declarator x

The first fix I could think of is to parameterize all undeclared variables in the template declaration.  Unfortunately it would not let me pass in s.b as an aliased argument to the mixin.  I finally got the following code to compile:

template Foo(alias y, alias z, alias q, int select: 1)
{
double x = y + z - q ;
}
template Foo(alias y, alias z, alias q, int select: 0)
{
double x = y + z + q ;
}

struct stuff {
double b = 5 ;
}

int main(char[][] args)
{
stuff s ;
double x, y = 2.0, z = 3.0, q = 5.0 ;
mixin Foo!(y, z, q ,1) ;
printf("x = %g\n", x) ;
return(0) ;
}

However, when I run it,  I get the following:
x = nan

I am trying to get the behavior of a conditional macro,  where I can make a compile time decision about what code is actually inserted during the mixin instantiation.  So far I have failed to duplicate the desired behaviour. I would appreciate a clarification on how mixin's are interpreted.

-Matt

It seems like a bug.

Workaround if you wish:

template Foo(int select: 1)
{
  double x = y + z - s.b ;
}
template Foo(int select: 0)
{
  double x = y + z + s.b ;
}

struct stuff {
  double b = 5 ;
}

int main(char[][] args)
{
  stuff s ;
  double y = 2.0, z = 3.0 ;
  mixin Foo!(1) ;
  writef("x = %g\n", x) ;
  return(0) ;
}

Andrew.



<mattcbro@earthlink.net> wrote in message news:d4eq7o$2nhj$1@digitaldaemon.com...
>I am baffled about how to use mixins.  In the online documentation for mixin's
> there is the following statement:
>
> " Unlike a template instantiation, a template mixin's body is evaluated
> within
> the scope where the mixin appears, not where the template declaration is
> defined. It is analogous to cutting and pasting the body of the template
> into
> the location of the mixin. It is useful for injecting parameterized
> 'boilerplate' code, as well as for creating templated nested functions,
> which is
> not possible with template instantiations."
>
> Unfortunately, the current version of the compiler does not evalute the
> mixin
> where it is declared per se,  in that all declarations local to the
> location of
> the mixin, but not visible to the template declaration cause compile
> errors.
> For example consider the following code, which attempts to select one of
> two
> possible computations:
>
> template Foo(int select: 1) {
> x = y + z - s.b ;
> }
> template Foo(int select: 0) {
> x = y + z + s.b ;
> }
>
> struct stuff {
> double b = 5 ;
> }
>
> int main(char[][] args)
> {
> stuff s ;
> double x, y = 2.0, z = 3.0 ;
> mixin Foo!(1) ;
> printf("x = %g\n", x) ;
> return(0) ;
> }
>
> This fails to compile with the error
> mixintest.d(2): no identifier for declarator x
> mixintest.d(5): no identifier for declarator x
>
> The first fix I could think of is to parameterize all undeclared variables
> in
> the template declaration.  Unfortunately it would not let me pass in s.b
> as an
> aliased argument to the mixin.  I finally got the following code to
> compile:
>
> template Foo(alias y, alias z, alias q, int select: 1)
> {
> double x = y + z - q ;
> }
> template Foo(alias y, alias z, alias q, int select: 0)
> {
> double x = y + z + q ;
> }
>
> struct stuff {
> double b = 5 ;
> }
>
> int main(char[][] args)
> {
> stuff s ;
> double x, y = 2.0, z = 3.0, q = 5.0 ;
> mixin Foo!(y, z, q ,1) ;
> printf("x = %g\n", x) ;
> return(0) ;
> }
>
> However, when I run it,  I get the following:
> x = nan
>
> I am trying to get the behavior of a conditional macro,  where I can make
> a
> compile time decision about what code is actually inserted during the
> mixin
> instantiation.  So far I have failed to duplicate the desired behaviour. I
> would
> appreciate a clarification on how mixin's are interpreted.
>
> -Matt
>
>
> 

Yes thanks for the input.  Based on my hacking attempts it seems that the mixin templates fail if the left hand side of an assignment statement does not have fully specified type declarations.  Here are examples of code that fails

template Foo(T) {
T lhs ;
lhs.b = y + z ;
}

struct stuff {
double b  ;
}

int main(char[][] args)
{
double y = 2.0, z = 3.0 ;
mixin Foo!(stuff) ;
printf("b = %g\n", lhs.b) ;
return(0) ;
}

compile fails with mixintest.d(3): no identifier for declarator lhs.b

this fails

template Foo() {
double lhs = y + z ;
s.b = lhs ;
}

struct stuff {
double b  ;
}

int main(char[][] args)
{
double y = 2.0, z = 3.0 ;
stuff s ;
mixin Foo!() ;
printf("b = %g\n", s.b) ;
return(0) ;
}

this gives the compiler error
mixintest.d(3): no identifier for declarator s.b


this works:

template Foo() {
double lhs = y + z ;
}

struct stuff {
double b  ;
}

int main(char[][] args)
{
double y = 2.0, z = 3.0 ;
stuff s ;
mixin Foo!() ;
s.b = lhs ;
printf("b = %g\n", s.b) ;
return(0) ;
}

In article <d4es8i$2oo5$1@digitaldaemon.com>, Andrew Fedoniouk says...
>
>It seems like a bug.
>
>Workaround if you wish:
>
>template Foo(int select: 1)
>{
>  double x = y + z - s.b ;
>}
>template Foo(int select: 0)
>{
>  double x = y + z + s.b ;
>}
>
>struct stuff {
>  double b = 5 ;
>}
>
>int main(char[][] args)
>{
>  stuff s ;
>  double y = 2.0, z = 3.0 ;
>  mixin Foo!(1) ;
>  writef("x = %g\n", x) ;
>  return(0) ;
>}
>
>Andrew.
>
>
>
><mattcbro@earthlink.net> wrote in message news:d4eq7o$2nhj$1@digitaldaemon.com...
>>I am baffled about how to use mixins.  In the online documentation for mixin's
>> there is the following statement:
>>
>> " Unlike a template instantiation, a template mixin's body is evaluated
>> within
>> the scope where the mixin appears, not where the template declaration is
>> defined. It is analogous to cutting and pasting the body of the template
>> into
>> the location of the mixin. It is useful for injecting parameterized
>> 'boilerplate' code, as well as for creating templated nested functions,
>> which is
>> not possible with template instantiations."
>>
>> Unfortunately, the current version of the compiler does not evalute the
>> mixin
>> where it is declared per se,  in that all declarations local to the
>> location of
>> the mixin, but not visible to the template declaration cause compile
>> errors.
>> For example consider the following code, which attempts to select one of
>> two
>> possible computations:
>>
>> template Foo(int select: 1) {
>> x = y + z - s.b ;
>> }
>> template Foo(int select: 0) {
>> x = y + z + s.b ;
>> }
>>
>> struct stuff {
>> double b = 5 ;
>> }
>>
>> int main(char[][] args)
>> {
>> stuff s ;
>> double x, y = 2.0, z = 3.0 ;
>> mixin Foo!(1) ;
>> printf("x = %g\n", x) ;
>> return(0) ;
>> }
>>
>> This fails to compile with the error
>> mixintest.d(2): no identifier for declarator x
>> mixintest.d(5): no identifier for declarator x
>>
>> The first fix I could think of is to parameterize all undeclared variables
>> in
>> the template declaration.  Unfortunately it would not let me pass in s.b
>> as an
>> aliased argument to the mixin.  I finally got the following code to
>> compile:
>>
>> template Foo(alias y, alias z, alias q, int select: 1)
>> {
>> double x = y + z - q ;
>> }
>> template Foo(alias y, alias z, alias q, int select: 0)
>> {
>> double x = y + z + q ;
>> }
>>
>> struct stuff {
>> double b = 5 ;
>> }
>>
>> int main(char[][] args)
>> {
>> stuff s ;
>> double x, y = 2.0, z = 3.0, q = 5.0 ;
>> mixin Foo!(y, z, q ,1) ;
>> printf("x = %g\n", x) ;
>> return(0) ;
>> }
>>
>> However, when I run it,  I get the following:
>> x = nan
>>
>> I am trying to get the behavior of a conditional macro,  where I can make
>> a
>> compile time decision about what code is actually inserted during the
>> mixin
>> instantiation.  So far I have failed to duplicate the desired behaviour. I
>> would
>> appreciate a clarification on how mixin's are interpreted.
>>
>> -Matt
>>
>>
>> 
>
>