template for function pointer

[firegun9@yahoo.com.tw]

Hello everyone,
here is my program:

///////////////////////////////
#include <iostream>
using namespace std;

void multi(double* arrayPtr, int len){
for(int i=0; i<len; i++)
*(arrayPtr+i)*= 2;
}
typedef void (*p2f) (double* a, int b);
//////////////////////////////
template <class T>
class bf{
private:
int type;
T var;
public:
bf(int i, T value){ type=i; var=value;}
void eval(double* dPtr, int len);
};

template <class T>
void bf<T>::eval(dou ble* dPtr, int len){
switch(type){
case 0:
for(int i=0; i<len; i++)
*(dPtr+i)=var;
break;
case 1:
for(int i=0; i<len; i++)
*(dPtr+i)=*(var +i);
break;
case 2:
var(dPtr, len);
break;
}
}

void main(){
double a=0.0;
bf<double> a1(0,a);

double b[3]={0.0, 1.0, 2.0};
bf<double*> a2(1,b);

bf<p2f> a3(2,multi);

double temp[3]={5.0, 5.0, 5.0};
a1.eval(temp,3) ;
for(int i=0;i<3;i++)
cout<<temp[i]<<endl;
}

I have a bf class with a member variable "var" whose type is defined as
a template. In bf's constructor, I use (int type) to record what type
var is defined.
The main purpose of bf is to retrieve a double array, then modified it
using the var member.
By different types of var, it does different modifies to the input
array.

There are 3 conditions:
If var is a double( it can be known from "type" variable), it fill the
incoming array with the double value.
If var is a double array, it copies its value one by one into the
incoming array.
If var is a function pointer, it takes the incoming array as the
argument.

There's no problem untill I call bf::eval() in main();
It seems like that if var is a double then the code "var(dPtr, len);"
is wrong. In fact, it is wrong. But that is in case 2. Whenever the var
is not a function pointer, it will never go into case 2 in the switch
in eval();
The familiar situation also happens when the var is set to a function
pointer. The error was found in case 0, in which var is treated as a
double.

I just wanna control the branch call by myself.
Is there any solution?
Thanks

[Thomas Wintschel]

Re: template for function pointer


<firegun9@yahoo .com.tw> wrote in message
news:1104616356 .358972.83090@z 14g2000cwz.goog legroups.com...[color=blue]
> Hello everyone,
> here is my program:
>
> ///////////////////////////////
> #include <iostream>
> using namespace std;
>
> void multi(double* arrayPtr, int len){
> for(int i=0; i<len; i++)
> *(arrayPtr+i)*= 2;
> }
> typedef void (*p2f) (double* a, int b);
> //////////////////////////////
> template <class T>
> class bf{
> private:
> int type;
> T var;
> public:
> bf(int i, T value){ type=i; var=value;}
> void eval(double* dPtr, int len);
> };
>[/color]

Generally, you use a template when you want to apply the same code
structure to different types. Your problem here is that you are trying
to apply different code depending on the type which fails because you
cannot provide a T that has all the necessary properties.
[color=blue]
> template <class T>
> void bf<T>::eval(dou ble* dPtr, int len){
> switch(type){
> case 0:
> for(int i=0; i<len; i++)
> *(dPtr+i)=var;[/color]

In order for this line to compile, it must be possible to convert T to a
double.
[color=blue]
> break;
> case 1:
> for(int i=0; i<len; i++)
> *(dPtr+i)=*(var +i);[/color]

In order for this line to compile, it must be possible to convert T to a
double*.
[color=blue]
> break;
> case 2:
> var(dPtr, len);[/color]

In order for this line to compile,
T must have a constructor that takes a double* and an int.
[color=blue]
> break;
> }
> }
>[/color]

So, if you want the preceding method to compile, you need to use a T for
which the following statements are all valid:

T t1;
double d = t1;
double* p = t1;
T t2(p, 1);
[color=blue]
> void main(){
> double a=0.0;
> bf<double> a1(0,a);
>
> double b[3]={0.0, 1.0, 2.0};
> bf<double*> a2(1,b);
>
> bf<p2f> a3(2,multi);
>
> double temp[3]={5.0, 5.0, 5.0};
> a1.eval(temp,3) ;
> for(int i=0;i<3;i++)
> cout<<temp[i]<<endl;
> }
>
> I have a bf class with a member variable "var" whose type is defined[/color]
as[color=blue]
> a template. In bf's constructor, I use (int type) to record what type
> var is defined.
> The main purpose of bf is to retrieve a double array, then modified it
> using the var member.
> By different types of var, it does different modifies to the input
> array.
>
> There are 3 conditions:
> If var is a double( it can be known from "type" variable), it fill the
> incoming array with the double value.
> If var is a double array, it copies its value one by one into the
> incoming array.
> If var is a function pointer, it takes the incoming array as the
> argument.
>
> There's no problem untill I call bf::eval() in main();
> It seems like that if var is a double then the code "var(dPtr, len);"
> is wrong. In fact, it is wrong. But that is in case 2. Whenever the[/color]
var[color=blue]
> is not a function pointer, it will never go into case 2 in the switch
> in eval();
> The familiar situation also happens when the var is set to a function
> pointer. The error was found in case 0, in which var is treated as a
> double.
>
> I just wanna control the branch call by myself.
> Is there any solution?
> Thanks
>[/color]

I will ignore other obvious, dangerous errors in your code since I think
you are really misguided in trying to do what you are doing.

Some advice:

First read some more about C++ and object-oriented programming. You are
trying to treat numerical, pointer and function pointer types like they
are the same thing when they are very different creatures.

Read much more about templates. In particular, take a look at STL type
requirements, which specify the properties a type must have in order to
be used with a particular template. A vector, for instance, requires
that T be 'Assignable' meaning that you can write:
T a;
T b = a;

Perhaps if you state what higher level problem you are trying to solve
you will get some useful advice. In the meantime, remember that
templates will only work with types that have something in common.

Merry New Year
Tom

[Bruce Trask]

Re: template for function pointer

Hi Thomas,

The problem you are seeing is that the compiler is trying to compile all
three branches of the case statement *for each type of the template
parameter* and all those branches are not valid of all the template
parameters you are using.

I am going to throw some code at you that perhaps solves you requirements in
different way. Have a look at it. It probably introduces some new concepts
but these concepts lend themselve more correctly to solving the type of
problem you are trying to solve.

///////////////////////////////////////

#include <vector>
#include <algorithm>
#include <iterator>
#include <iostream>

using namespace std;

double doubleArray[] = { 1.0, 2.0, 3.0 };
const size_t sz = 3;


double doubleArrayTwo[] = { 5.0, 6.0, 7.0 };
const size_t sz2 = 3;

double func(const double elem)
{
return elem*2;
}
int main()
{
vector<double> myDoubleVec(&do ubleArray[0],
&doubleArray[3]);

// case 1
double val = 5;

fill(myDoubleVe c.begin(),
myDoubleVec.end (),
5);

// print the result
copy(myDoubleVe c.begin(),
myDoubleVec.end (),
ostream_iterato r<double>(cout, "\n"));

// case 2

copy(&doubleArr ayTwo[0],
&doubleArray Two[sz2],
myDoubleVec.beg in());

// print the result
copy(myDoubleVe c.begin(),
myDoubleVec.end (),
ostream_iterato r<double>(cout, "\n"));

// case 3
transform(myDou bleVec.begin(),
myDoubleVec.end (),
myDoubleVec. begin(),
func);

copy(myDoubleVe c.begin(),
myDoubleVec.end (),
ostream_iterato r<double>(cout, "\n"));

return 0;
}
///////////////////////////////////////////////////////////////////////
Let me know how it goes.

Regards,
Bruce


<firegun9@yahoo .com.tw> wrote in message
news:1104616356 .358972.83090@z 14g2000cwz.goog legroups.com...[color=blue]
> Hello everyone,
> here is my program:
>
> ///////////////////////////////
> #include <iostream>
> using namespace std;
>
> void multi(double* arrayPtr, int len){
> for(int i=0; i<len; i++)
> *(arrayPtr+i)*= 2;
> }
> typedef void (*p2f) (double* a, int b);
> //////////////////////////////
> template <class T>
> class bf{
> private:
> int type;
> T var;
> public:
> bf(int i, T value){ type=i; var=value;}
> void eval(double* dPtr, int len);
> };
>
> template <class T>
> void bf<T>::eval(dou ble* dPtr, int len){
> switch(type){
> case 0:
> for(int i=0; i<len; i++)
> *(dPtr+i)=var;
> break;
> case 1:
> for(int i=0; i<len; i++)
> *(dPtr+i)=*(var +i);
> break;
> case 2:
> var(dPtr, len);
> break;
> }
> }
>
> void main(){
> double a=0.0;
> bf<double> a1(0,a);
>
> double b[3]={0.0, 1.0, 2.0};
> bf<double*> a2(1,b);
>
> bf<p2f> a3(2,multi);
>
> double temp[3]={5.0, 5.0, 5.0};
> a1.eval(temp,3) ;
> for(int i=0;i<3;i++)
> cout<<temp[i]<<endl;
> }
>
> I have a bf class with a member variable "var" whose type is defined as
> a template. In bf's constructor, I use (int type) to record what type
> var is defined.
> The main purpose of bf is to retrieve a double array, then modified it
> using the var member.
> By different types of var, it does different modifies to the input
> array.
>
> There are 3 conditions:
> If var is a double( it can be known from "type" variable), it fill the
> incoming array with the double value.
> If var is a double array, it copies its value one by one into the
> incoming array.
> If var is a function pointer, it takes the incoming array as the
> argument.
>
> There's no problem untill I call bf::eval() in main();
> It seems like that if var is a double then the code "var(dPtr, len);"
> is wrong. In fact, it is wrong. But that is in case 2. Whenever the var
> is not a function pointer, it will never go into case 2 in the switch
> in eval();
> The familiar situation also happens when the var is set to a function
> pointer. The error was found in case 0, in which var is treated as a
> double.
>
> I just wanna control the branch call by myself.
> Is there any solution?
> Thanks
>[/color]

[Bruce Trask]

Re: template for function pointer

Sorry, my response was not to Thomas but to the original poster.


"Bruce Trask" <b_trask@yahoo. com> wrote in message
news:AuKBd.1846 9$152.17240@trn dny01...[color=blue]
> Hi Thomas,
>
> The problem you are seeing is that the compiler is trying to compile all
> three branches of the case statement *for each type of the template
> parameter* and all those branches are not valid of all the template
> parameters you are using.
>
> I am going to throw some code at you that perhaps solves you requirements[/color]
in[color=blue]
> different way. Have a look at it. It probably introduces some new[/color]
concepts[color=blue]
> but these concepts lend themselve more correctly to solving the type of
> problem you are trying to solve.
>
> ///////////////////////////////////////
>
> #include <vector>
> #include <algorithm>
> #include <iterator>
> #include <iostream>
>
> using namespace std;
>
> double doubleArray[] = { 1.0, 2.0, 3.0 };
> const size_t sz = 3;
>
>
> double doubleArrayTwo[] = { 5.0, 6.0, 7.0 };
> const size_t sz2 = 3;
>
> double func(const double elem)
> {
> return elem*2;
> }
> int main()
> {
> vector<double> myDoubleVec(&do ubleArray[0],
> &doubleArray[3]);
>
> // case 1
> double val = 5;
>
> fill(myDoubleVe c.begin(),
> myDoubleVec.end (),
> 5);
>
> // print the result
> copy(myDoubleVe c.begin(),
> myDoubleVec.end (),
> ostream_iterato r<double>(cout, "\n"));
>
> // case 2
>
> copy(&doubleArr ayTwo[0],
> &doubleArray Two[sz2],
> myDoubleVec.beg in());
>
> // print the result
> copy(myDoubleVe c.begin(),
> myDoubleVec.end (),
> ostream_iterato r<double>(cout, "\n"));
>
> // case 3
> transform(myDou bleVec.begin(),
> myDoubleVec.end (),
> myDoubleVec. begin(),
> func);
>
> copy(myDoubleVe c.begin(),
> myDoubleVec.end (),
> ostream_iterato r<double>(cout, "\n"));
>
> return 0;
> }
> ///////////////////////////////////////////////////////////////////////
> Let me know how it goes.
>
> Regards,
> Bruce
>
>
> <firegun9@yahoo .com.tw> wrote in message
> news:1104616356 .358972.83090@z 14g2000cwz.goog legroups.com...[color=green]
> > Hello everyone,
> > here is my program:
> >
> > ///////////////////////////////
> > #include <iostream>
> > using namespace std;
> >
> > void multi(double* arrayPtr, int len){
> > for(int i=0; i<len; i++)
> > *(arrayPtr+i)*= 2;
> > }
> > typedef void (*p2f) (double* a, int b);
> > //////////////////////////////
> > template <class T>
> > class bf{
> > private:
> > int type;
> > T var;
> > public:
> > bf(int i, T value){ type=i; var=value;}
> > void eval(double* dPtr, int len);
> > };
> >
> > template <class T>
> > void bf<T>::eval(dou ble* dPtr, int len){
> > switch(type){
> > case 0:
> > for(int i=0; i<len; i++)
> > *(dPtr+i)=var;
> > break;
> > case 1:
> > for(int i=0; i<len; i++)
> > *(dPtr+i)=*(var +i);
> > break;
> > case 2:
> > var(dPtr, len);
> > break;
> > }
> > }
> >
> > void main(){
> > double a=0.0;
> > bf<double> a1(0,a);
> >
> > double b[3]={0.0, 1.0, 2.0};
> > bf<double*> a2(1,b);
> >
> > bf<p2f> a3(2,multi);
> >
> > double temp[3]={5.0, 5.0, 5.0};
> > a1.eval(temp,3) ;
> > for(int i=0;i<3;i++)
> > cout<<temp[i]<<endl;
> > }
> >
> > I have a bf class with a member variable "var" whose type is defined as
> > a template. In bf's constructor, I use (int type) to record what type
> > var is defined.
> > The main purpose of bf is to retrieve a double array, then modified it
> > using the var member.
> > By different types of var, it does different modifies to the input
> > array.
> >
> > There are 3 conditions:
> > If var is a double( it can be known from "type" variable), it fill the
> > incoming array with the double value.
> > If var is a double array, it copies its value one by one into the
> > incoming array.
> > If var is a function pointer, it takes the incoming array as the
> > argument.
> >
> > There's no problem untill I call bf::eval() in main();
> > It seems like that if var is a double then the code "var(dPtr, len);"
> > is wrong. In fact, it is wrong. But that is in case 2. Whenever the var
> > is not a function pointer, it will never go into case 2 in the switch
> > in eval();
> > The familiar situation also happens when the var is set to a function
> > pointer. The error was found in case 0, in which var is treated as a
> > double.
> >
> > I just wanna control the branch call by myself.
> > Is there any solution?
> > Thanks
> >[/color]
>
>
>[/color]

[Bruce Trask]

Re: template for function pointer

Aside from taking a different approach with the STL, you had a question as
to how to make the compiler avoid some branches in the code. The answer to
that would be to provide template specializations for the eval member
function. This gives you what amounts to a compile time switch. See the
code below to see how this works. It has the added benefit of failing at
link time if anyone uses a type that you have not provided a specialization
for.

#include <iostream>
using namespace std;

void multi(double* arrayPtr, int len){
for(int i=0; i<len; i++)
*(arrayPtr+i)*= 2;
}
typedef void (*p2f) (double* a, int b);
//////////////////////////////
template <class T>
class bf{
private:
int type;
T var;
public:
bf(int i, T value){ type=i; var=value;}
void eval(double* dPtr, int len);
};

template <>
void bf<double>::eva l(double* dPtr, int len){
cout << "1" << endl;
for(int i=0; i<len; i++)
*(dPtr+i)=var;
}

template <>
void bf<double*>::ev al(double* dPtr, int len){
cout << "2" << endl;
for(int i=0; i<len; i++)
*(dPtr+i)=*(var +i);
}

template <>
void bf<p2f>::eval(d ouble* dPtr, int len){
cout << "3" << endl;
var(dPtr, len);
}

int main(){
double a=0.0;
bf<double> a1(0,a);

double b[3]={0.0, 1.0, 2.0};
bf<double*> a2(1,b);

bf<p2f> a3(2,multi);

double temp[3]={5.0, 5.0, 5.0};
cout << "calling eval" << endl;
a1.eval(temp,3) ;
for(int i=0;i<3;i++)
cout<<temp[i]<<endl;
}

Hope that helps,
Bruce

"Bruce Trask" <b_trask@yahoo. com> wrote in message
news:6wKBd.1042 7$sh5.8053@trnd ny08...[color=blue]
> Sorry, my response was not to Thomas but to the original poster.
>
>
> "Bruce Trask" <b_trask@yahoo. com> wrote in message
> news:AuKBd.1846 9$152.17240@trn dny01...[color=green]
> > Hi Thomas,
> >
> > The problem you are seeing is that the compiler is trying to compile all
> > three branches of the case statement *for each type of the template
> > parameter* and all those branches are not valid of all the template
> > parameters you are using.
> >
> > I am going to throw some code at you that perhaps solves you[/color][/color]
requirements[color=blue]
> in[color=green]
> > different way. Have a look at it. It probably introduces some new[/color]
> concepts[color=green]
> > but these concepts lend themselve more correctly to solving the type of
> > problem you are trying to solve.
> >
> > ///////////////////////////////////////
> >
> > #include <vector>
> > #include <algorithm>
> > #include <iterator>
> > #include <iostream>
> >
> > using namespace std;
> >
> > double doubleArray[] = { 1.0, 2.0, 3.0 };
> > const size_t sz = 3;
> >
> >
> > double doubleArrayTwo[] = { 5.0, 6.0, 7.0 };
> > const size_t sz2 = 3;
> >
> > double func(const double elem)
> > {
> > return elem*2;
> > }
> > int main()
> > {
> > vector<double> myDoubleVec(&do ubleArray[0],
> > &doubleArray[3]);
> >
> > // case 1
> > double val = 5;
> >
> > fill(myDoubleVe c.begin(),
> > myDoubleVec.end (),
> > 5);
> >
> > // print the result
> > copy(myDoubleVe c.begin(),
> > myDoubleVec.end (),
> > ostream_iterato r<double>(cout, "\n"));
> >
> > // case 2
> >
> > copy(&doubleArr ayTwo[0],
> > &doubleArray Two[sz2],
> > myDoubleVec.beg in());
> >
> > // print the result
> > copy(myDoubleVe c.begin(),
> > myDoubleVec.end (),
> > ostream_iterato r<double>(cout, "\n"));
> >
> > // case 3
> > transform(myDou bleVec.begin(),
> > myDoubleVec.end (),
> > myDoubleVec. begin(),
> > func);
> >
> > copy(myDoubleVe c.begin(),
> > myDoubleVec.end (),
> > ostream_iterato r<double>(cout, "\n"));
> >
> > return 0;
> > }
> > ///////////////////////////////////////////////////////////////////////
> > Let me know how it goes.
> >
> > Regards,
> > Bruce
> >
> >
> > <firegun9@yahoo .com.tw> wrote in message
> > news:1104616356 .358972.83090@z 14g2000cwz.goog legroups.com...[color=darkred]
> > > Hello everyone,
> > > here is my program:
> > >
> > > ///////////////////////////////
> > > #include <iostream>
> > > using namespace std;
> > >
> > > void multi(double* arrayPtr, int len){
> > > for(int i=0; i<len; i++)
> > > *(arrayPtr+i)*= 2;
> > > }
> > > typedef void (*p2f) (double* a, int b);
> > > //////////////////////////////
> > > template <class T>
> > > class bf{
> > > private:
> > > int type;
> > > T var;
> > > public:
> > > bf(int i, T value){ type=i; var=value;}
> > > void eval(double* dPtr, int len);
> > > };
> > >
> > > template <class T>
> > > void bf<T>::eval(dou ble* dPtr, int len){
> > > switch(type){
> > > case 0:
> > > for(int i=0; i<len; i++)
> > > *(dPtr+i)=var;
> > > break;
> > > case 1:
> > > for(int i=0; i<len; i++)
> > > *(dPtr+i)=*(var +i);
> > > break;
> > > case 2:
> > > var(dPtr, len);
> > > break;
> > > }
> > > }
> > >
> > > void main(){
> > > double a=0.0;
> > > bf<double> a1(0,a);
> > >
> > > double b[3]={0.0, 1.0, 2.0};
> > > bf<double*> a2(1,b);
> > >
> > > bf<p2f> a3(2,multi);
> > >
> > > double temp[3]={5.0, 5.0, 5.0};
> > > a1.eval(temp,3) ;
> > > for(int i=0;i<3;i++)
> > > cout<<temp[i]<<endl;
> > > }
> > >
> > > I have a bf class with a member variable "var" whose type is defined[/color][/color][/color]
as[color=blue][color=green][color=darkred]
> > > a template. In bf's constructor, I use (int type) to record what type
> > > var is defined.
> > > The main purpose of bf is to retrieve a double array, then modified it
> > > using the var member.
> > > By different types of var, it does different modifies to the input
> > > array.
> > >
> > > There are 3 conditions:
> > > If var is a double( it can be known from "type" variable), it fill the
> > > incoming array with the double value.
> > > If var is a double array, it copies its value one by one into the
> > > incoming array.
> > > If var is a function pointer, it takes the incoming array as the
> > > argument.
> > >
> > > There's no problem untill I call bf::eval() in main();
> > > It seems like that if var is a double then the code "var(dPtr, len);"
> > > is wrong. In fact, it is wrong. But that is in case 2. Whenever the[/color][/color][/color]
var[color=blue][color=green][color=darkred]
> > > is not a function pointer, it will never go into case 2 in the switch
> > > in eval();
> > > The familiar situation also happens when the var is set to a function
> > > pointer. The error was found in case 0, in which var is treated as a
> > > double.
> > >
> > > I just wanna control the branch call by myself.
> > > Is there any solution?
> > > Thanks
> > >[/color]
> >
> >
> >[/color]
>
>[/color]