One line debug macro

Question

I'd like to create a C++ macro function to debug, and I'd like it to work like this:

int main(){
    int a = 3, b = 5, c = 7;
    string s = "<";

    print(a,s,b);
    print(a,s,b,s,c);
}

OUTPUT:
3 < 5
3 < 5 < 7

I've read a lot about variadic macros but anything I tried to code wouldnt work at all.

I thought about using lambda but didn't come up with algorithm to do it.

I need it like 1 line of code, cos it's just for debugging and more than that I could create a more complex function, but I guess this must be possible...

A macro cannot involve a runtime type such as std::string. Lambdas are a runtime thing too. Macros are for preprocessing, prior to any compilation, leave alone runtime. — cadaniluk
– cadaniluk, Commented Jul 31, 2016 at 20:27
what about char* ? or words that are not string, for example print(abc) where abc can be treated like #abc — Daniel
– Daniel, Commented Jul 31, 2016 at 20:29
If abc can be treated like #abc then it is something from the preprocessor. char* cannot do this either. int cannot be accessed form the preprocessor either. You could write some analyzer program to figure out the values of variables at a certain period of time within the execution of the program or just use a debugger or something that works at runtime. The C preprocessor has its limitations. m4 is a nice alternative that might come in handy here. — cadaniluk
– cadaniluk, Commented Jul 31, 2016 at 20:32

Leon · Accepted Answer · 2016-07-31 21:07:28Z

2

If you agree to replace the separator with '<<' (instead of ',') then the macro is very easy to define (note however that print is not a good name for a macro, that's why I renamed it appropriately):

#include <iostream>
#include <string>

#ifdef DEBUG
#define DEBUG_PRINT(x) std::cout << x << std::endl
#else
#define DEBUG_PRINT(x)
#endif

int main(){
    int a = 3, b = 5, c = 7;
    std::string s = "<";

    std::cout << "START" << std::endl;

    DEBUG_PRINT(a << s << b);
    DEBUG_PRINT(a << s << b << s << c);

    std::cout << "END" << std::endl;
    return 0;
}

Output:

$ g++ -DDEBUG main.cpp && ./a.out
START
3<5
3<5<7
END

$ g++ main.cpp && ./a.out
START
END

edited Jul 31, 2016 at 21:07

answered Jul 31, 2016 at 20:58

Leon

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Comments

Jarod42 · Accepted Answer · 2016-07-31 21:57:01Z

2

You may use variadic template instead of MACRO:

template <typename ... Ts>
void print(Ts&&... args)
{
    int dummy[] = {0, ((std::cout << args), 0)...};
    static_cast<void>(dummy); // avoid warning for unused variable
    std::cout << std::endl;
}

or in c++17

template <typename ... Ts>
void print(Ts&&... args)
{
    (std::cout << ... << args) << std::endl;
}

answered Jul 31, 2016 at 21:57

Jarod42

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Comments

Daniel · Accepted Answer · 2016-08-01 04:24:31Z

1

I almost achieved what I wanted (with 2 lines of code):

#define print(args...) {db,args; cerr<<endl;}
struct dbg{template<typename T> dbg& operator , (const T& v){cerr<<v<<" "; return *this; }} db;

now I can just print(1,2,"asjd") and it works fine

answered Aug 1, 2016 at 4:24

Daniel

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3 Comments

Jarod42 Over a year ago

You can even declare and define the struct inside the block since C++11 (and so remove the global).

Daniel Over a year ago

you mean I can declare the struct inside de macro? how?

Jarod42 Over a year ago

In fact you can declare local struct inside function, but NOT template method, so I was wrong. :/

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