Default parameter value in template function, depending on type

Question

Suppose I have this function to trim an std::string and decide to extend it, so that it removes not only spaces from the beginning and end, but also any character I pass in another string, like spaces, newlines and carriage returns.

std::string Trim ( const std::string &In_Original,
  const std::string &In_CharsToTrim = " \n\r" );

So basically, it would remove all characters present in In_CharsToTrim from the beginning and end of In_Original.

Now, I had to write this for both std::string and std::wstring. Since I find that absurd, I chose to make use of templates, and it works, but I am unable to pass a default value, because the std::wstring version must get L" \n\r" as the default value (note that little L, there).

I tried this:

template <typename Type> Type Trim ( const Type &In_String,
  const Type &In_TrimCharacters );

along with:

template std::string Trim ( const  std::string &In_String,
  const  std::string &In_TrimCharacters = " \n\r" );
template std::wstring Trim ( const  std::wstring &In_String,
  const  std::wstring &In_TrimCharacters = L" \n\r" );

But it didn't work. It doesn't even compile.

Currently, what I have, is a separate function for when you call without a second parameter, but this is completely the wrong approach:

template <typename Type> Type Trim ( const Type &In_String,
  const Type &In_TrimCharacters );
std::string Trim ( const std::string &In_String );
std::wstring Trim ( const std::wstring &In_String );

Then, in this simpler Trim function, I'm simply calling the complete version.

So, basically, what I'm asking is this...:

How can I pass a different default value depending on the template type? Basically, a specialization, in which the only change is the default parameter...

In this case, to pass std::string ( " \n\r" ) in one case, and std::wstring ( L" \n\r" ) in the other...

Or... is there another way to do what I'm trying to do, here?

T.C. · Accepted Answer · 2014-07-11 16:38:47Z

6

This is what I come up with:

template <typename T> struct default_trim_chars;

template<> struct default_trim_chars<std::string> { 
    static const char* value() { return " \n\r"; }
};

template<> struct default_trim_chars<std::wstring> { 
    static const wchar_t* value() { return L" \n\r"; }
};

template <typename Type> Type Trim ( const Type &In_String,
const Type &In_TrimCharacters = default_trim_chars<Type>::value()){
    /* ... */
}

You can also declare value as a data member instead of a function, but that would require constexpr and C++11.

answered Jul 11, 2014 at 16:38

T.C.

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1 Comment

Joao Pincho Over a year ago

This solution worked without changing compiler options, which was a must. Thanks!

David G · Accepted Answer · 2014-07-11 16:39:05Z

3

You can use a simple trait class and using its ::value as the default argument.

template<typename T>
struct type_sensitive_default_argument;

template<>
struct type_sensitive_default_argument<std::string>
{
     static constexpr char value[] = "\n\r";
};

template<>
struct type_sensitive_default_argument<std::wstring>
{
    static constexpr wchar_t value[] = L"\n\r";
};

Then you can do:

template<class Type,
         class Default = type_sensitive_default_argument<Type>>
Type Trim(const Type& In_String,
          const Type& In_TrimCharacters = Default::value);

answered Jul 11, 2014 at 16:39

David G

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2 Comments

Joao Pincho Over a year ago

For some reason, i am unable to compile that code. It does not recognize constexpr. I replaced it with simply const, and of course, it complained that i'm initializing a const member variable inside the class declaration. So i removed the initialization, and put it outside the declaration, and the i hit a brick wall. I got this: error: default template arguments may not be used in function templates without -std=c++11 or -std=gnu++11

Joao Pincho Over a year ago

I guess all of the above errors are because i'm still using C99... bummer. I'll try and see what the differences are, to see if it's possible to change the compiler settings, without much trouble..

Daniel Frey · Accepted Answer · 2014-07-11 18:50:43Z

1

In your case you might rely on specifying an initializer list of characters to do the conversion for you:

template <typename Type> Type Trim ( const Type &In_String,
  const Type &In_TrimCharacters = { ' ', '\r', '\n' } )
{
    // ...
}

answered Jul 11, 2014 at 18:50

Daniel Frey

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