Specializing function template for both std::string and char*

Question

As the title says I want to specialize a function template for both string and char pointer, so far I did this but I can not figure out passing the string parameters by reference.

#include <iostream>
#include <string.h>
template<typename T> void xxx(T param)
{
std::cout << "General : "<< sizeof(T)  << std::endl;
}

template<> void xxx<char*>(char* param)
{
std::cout << "Char ptr: "<< strlen(param) << std::endl;
}

template<> void xxx<const char* >(const char*  param)
{
std::cout << "Const Char ptr : "<< strlen(param)<<  std::endl;
}

template<> void xxx<const std::string & >(const std::string & param)
{
std::cout << "Const String : "<< param.size()<<  std::endl;
}

template<> void xxx<std::string >(std::string param)
{
std::cout << "String : "<< param.size()<<  std::endl;
}


int main()
{
        xxx("word");
        std::string aword("word");
        xxx(aword);

        std::string const cword("const word");
        xxx(cword);
}

Also template<> void xxx<const std::string & >(const std::string & param) thing just does not working.

If I rearranged the opriginal template to accept parameters as T& then the char * is required to be char * & which is not good for static text in code.

Please help !

Now it doesn't compile! You need to put <string.h> back, for strlen. — TonyK
– TonyK, Commented Jan 12, 2011 at 16:39

Konrad Rudolph · Accepted Answer · 2011-01-12 14:51:08Z

11

Doesn’t the following work?

template<>
void xxx<std::string>(std::string& param)
{
    std::cout << "String : "<< param.size()<<  std::endl;
}

And the same for const std::string?

That said, don’t specialize a function template if you have the choice (and you usually do!). Instead, just overload the function:

void xxx(std::string& param)
{
    std::cout << "String : "<< param.size()<<  std::endl;
}

Notice, this is not a template. In 99% of the cases, this is fine.

(Something else, C++ doesn’t have a header <string.h> except for backwards compatibility to C. The C-string header in C++ is called <cstring> (note the leading c) but from your code it look as though you actually mean the header <string> (no leading c).)

edited Jan 12, 2011 at 14:51

answered Jan 12, 2011 at 14:33

Konrad Rudolph

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5 Comments

ali_bahoo Over a year ago

Ah sorry for the <string.h>. I will fix it.

stefaanv Over a year ago

A link would be nice as to why it shouldn't be specialized, so here it is: gotw.ca/publications/mill17.htm

Konrad Rudolph Over a year ago

@stefaanv: nice, will include.

stefaanv Over a year ago

@sad_man: no, it's in Konrads answer and to be honest, I wasn't aware of this before...

TonyK Over a year ago

Have you compiled this? g++ 4.4.0 doesn't accept it. It gives the error message: error: template-id 'xxx<std::string>' for 'void xxx(std::string&)' does not match any template declaration

TonyK · Accepted Answer · 2011-01-12 17:00:19Z

0

Here is a distillation of what I find surprising:

#include <iostream>

template<typename T> void f(T param) { std::cout << "General" << std::endl ; }
template<> void f(int& param) { std::cout << "int&" << std::endl ; }

int main() {
  float x ; f (x) ;
  int y ; f (y) ;
  int& z = y ; f (z) ;
}

This prints "General" 3 times. The first time (float) is expected, the third time (int&) is a surprise. Why doesn't this work?

answered Jan 12, 2011 at 17:00

TonyK

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1 Comment

ali_bahoo Over a year ago

This is the problem I have with std::string&. There is something wrong with passing referenced types in a template. I wish somebody explains it.

Gustavo V · Accepted Answer · 2011-01-12 17:35:01Z

0

is really risky try to coding using compiler-based-type-conversions

one thing is the use of a template based, and other is the use of polymorph with different types.

depend of the compiler you can get different behaviors.

answered Jan 12, 2011 at 17:35

Gustavo V

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