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c++debuggingmacrosvariadic-functions
Nikhil Garg
Nikhil Garg

Reputation: 4014

Debug macro for C++ with variable arguments without the format string

Is it possible to write a macro which can take in a variable number of arguments and expands like this :

quickdebug(a)   ->  cout << #a ": " << a;
quickdebug(a,b) ->  cout << #a ": " << a << #b ": "<< b;

etc

If not, is it possible for me to at least print all the arguments without giving format strings. e.g

quickdebug2(a)   -> cout << a ;
quickdebug2(a,b) -> cout << a << " " << b ;

etc

For example in java I can write a function which provides me similar functionality: 

void debug(Object...args) 
{
  System.out.println(Arrays.deepToString(args));
}

Upvotes: 3

Views: 2114

Answers (2)

Robert
Robert

Reputation: 2389

By using a class that overrides , operator:

class VariadicToOutputStream
{
public:
    VariadicToOutputStream(std::ostream& s, const std::string& separator = " ") : m_stream(s), m_hasEntries(false), m_separator(separator) {}
    template<typename ObjectType>
    VariadicToOutputStream& operator , (const ObjectType& v)
    {
        if (m_hasEntries) m_stream << m_separator;
        m_stream << v;
        m_hasEntries=true;
        return *this;
    }
    ~VariadicToOutputStream()
    {
        m_stream << std::endl;
    }

private:
    std::ostream& m_stream;
    bool m_hasEntries;
    std::string m_separator;
};

You can write for instance:

VariadicToOutputStream(std::cout) , 1, 0.5f, "a string";

This can then be wrapped with a preprocessor macro:

#define VARIADIC_TO_STDOUT(...)     VariadicToOutputStream(std::cout),__VA_ARGS__;

So you can write:

VARIADIC_TO_STDOUT(1, 0.5f, "a string");

It would be easy to add f.i. separator strings to be used between arguments.

Edit: I just added a default space as separator string.

Upvotes: 7

Tommy Andersen
Tommy Andersen

Reputation: 7230

It is possible to make a macro that is variadic thus taking a variable amount of arguments. The syntax is similar to that of a function:

#define quickdebug(...) functiontocall("test", __VA_ARGS__)

Any argument listed after the last named argument in the argument list will be listed in __VA_ARGS__ including any seperating comma.

So: quickdebug(1, 2, "123", 4.5) becomes functioncall("test", 1, 2 , "123", 4.5)

However at some point you need to use these arguments, and here it can become extremely difficult if you don't have a format string, or something else indicating the type of the arguments.

The problem is that when reading variables from a variable arguments list, you need to know the type of the argument, or at least its size. If I were you I would choose a different approach.

You can read more about variadic macros here: http://gcc.gnu.org/onlinedocs/cpp/Variadic-Macros.html

Upvotes: 1

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