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Rust's dbg! is one of those ideas that felt obvious and beautiful once I saw it. Apparently it's inspired by Haskell.[0] I'd love to see it officially picked u

• lock-free 6y ago
  I've been using it in C/C++ projects going back at least 15 years, I thought it was commonplace for every language
  • dagmx 6y ago
    Do you have an example of using it in C or C++?
    • lock-free 6y ago
      https://stackoverflow.com/questions/1644868/define-macro-for-debug-printing-in-c https://stackoverflow.com/questions/1644868/define-macro-for...
      • steveklabnik 6y ago
        There's a few significant differences here. This is a printf that only happens in debug mode; dbg is something that takes an expression, returns that expression, and will print out its filename, line number, and a pretty-printed version of the value. For an example of how this plays out, I had some code: let x = y + z; I wasn't sure what y was, so I was able to let x = dbg!(y) + z; this is much easier to insert into (and remove from) existing code. Plus, you don't need to write the formatting code. Basically, what you posted is in the same genre of thing, but is very different in a lot of meaningful ways.
        • digikata 6y ago
          You can use dbg inline within an expression!? How did I miss that?
          • steveklabnik 6y ago
            You can even use it inside itself. I've written let c = dbg!(dbg!(a) + dbg!(b)); one time...
            • jrimbault 6y ago
              I've written a 3 nested dbg! line for a complex boolean expression... I should have just written a truth table...
              • brodie 6y ago
                Maybe you should write a truth table/source code macro!
        • CJefferson 6y ago
          After I saw this in Rust, I had to implement it in C++: #include <iostream> #include <string> template<typename T> T print_dbg(T val, std::string exp, std::string file, int line) { std::cerr << exp << " = " << val << " @ " << file << ":" << line << "\n"; return val; } #define dbg(x) print_dbg(x,#x,__FILE__,__LINE__)
          • petschge 6y ago
            Close. Applying that to double c = dbg(dbg(a) + dbg(b)); gives a = 1 @ debug.cxx:16 b = 2 @ debug.cxx:16 dbg(a) + dbg(b) = 3 @ debug.cxx:16 which has extra dbg()s on the third line. If I could I would probably want to write double dbg(c) = dbg(a) + dbg(b); but that of course doesn't work.
      • dagmx 6y ago
        Like Steve said, what you're talking about is quite a bit different than what the rust dbg macro does.
        • CyberDildonics 6y ago
          It isn't really that different, it can just be used in inline expressions. Having the macro do nothing or just evaluating to the expression in debug mode is trivial, since a debug symbol would be defined.