• mattcbro earthlink.net (62/62) Apr 23 2005 I am baffled about how to use mixins. In the online documentation for m...
• Andrew Fedoniouk (24/101) Apr 23 2005 It seems like a bug.
• mattcbro earthlink.net (53/169) Apr 23 2005 Yes thanks for the input. Based on my hacking attempts it seems that th...

mattcbro earthlink.net writes:

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

"Andrew Fedoniouk" <news terrainformatica.com> writes:

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


 

mattcbro earthlink.net writes:

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