Parsing error?

Hi all, I'm hoping someone that actually knows c++ can help me with this; I've never programmed in c++ but my current task is porting a c++ program from unix to windows (dos command line really) and while I've gotten to where I'm compiling it in dos, I get the error "parse error before 'if'" in one of the header files. The code looks like this:

Code:

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#ifndef __TRACE__
#define __TRACE__


#include <iostream>


/* global Variable of Trace-Level */
extern int TraceLevel;

#ifndef NDEBUG
#define Trace(Level, Output) \
        if (Level <= TraceLevel) { //parsing error on this line
    cout << Output << endl << flush;
  }

#define TraceIf( Condition, Level, Output) \
  if ((Condition) && (Level <= TraceLevel) ) {
    cout << Output << endl << flush;
  }
...

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and it's being called from this method in another header file:

Code:

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CPairDict<T1,T2>::~CPairDict() {
  Trace(2, "Enter CPairDict<T1,T2>::~CPairDict()");
  clear();
  Trace(2, "Leave CPairDict<T1,T2>::~CPairDict()");
}

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As I said, I barely know any c++, so whatever is wrong with this code eludes me (it compiles and runs fine under both unix and cygwin) since I didn't write it to begin with and barely know what it does. Any help would be really appreciated.

I think the problem is that since you are using a macro, you need to have the backslashes to escape the newline sequences.

The correct version is probably:

Code:

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#define Trace(Level, Output) \
        if (Level <= TraceLevel) {\
                cout << Output << endl << flush;\
        }

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Frankly, it may be better to use function templates:

Code:

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template <typename T>
inline void Trace(int Level, const T& Output) {
        if (Level <= TraceLevel) {
                std::cout << Output << std::endl;
        }
}

template <typename T>
inline void TraceIf(bool Condition, int Level, const T& Output) {
        if (Condition && (Level <= TraceLevel)) {
                std::cout << Output << std::endl;
        }
}

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Note that there is no need to use std::flush after std::endl since flush is incorporated into endl.

The thing is, the program had the back slashes you mentioned in it originally, but with them in, it not only gives the parse error but claims to have stray \'s in the program (which is why I removed them before).

Quote:

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The thing is, the program had the back slashes you mentioned in it originally, but with them in, it not only gives the parse error but claims to have stray \'s in the program (which is why I removed them before).

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That's strange. Do you still get the same problem with the code example I posted? Anyway, you might as well switch from using macros to using (inline) function templates.

Well, there's a couple things you should do then:
a) Put the backslashes back in, and tell us specifically what errors come up (and where). This way, we can figure out what the *real* problem is, and you'll understand better what's going on.
b) Try laserlight's solution; inline/templated functions are usually preferred over macros in C++ (many reasons; you can probably find something in the FAQ).

> the parse error but claims to have stray \'s in the program
Make sure it's absolutely the last character on the line.
Otherwise, it will be a stray \

I think it was a problem with the macros, when I use the inline templates it seems to stop giving the error on that function (I just get it on the next macro after those), so I'll definitely try converting them all. Thank you both for your help!

Quote:

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and it's being called from this method in another header file:

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nitpick warning! The following is me being annoying.

That is not a method. It's called destructor. Virtual member functions can be called methods. But that's about it as far as the term can be applied to C++ constructs. It's even shun by some authors. Languages like java and visual basic use the term method to describe a function member of the class. C++ doesn't.

Stroustrup C++ glossary

Quote:

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As I said, I barely know any c++

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And that is why I'm considering myself annoying :)

Quote:

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That is not a method. It's called destructor. Virtual member functions can be called methods.

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Well, that would make the destructor of a class intended as a base class a method, wouldnt it? :p

LOL. Ok

I noticed that you appeared to be using

Code:

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cout << ...;

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without declaring the std namespace

Code:

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using namespace std;

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or using

Code:

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std::cout <<...;

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I'm not sure if that's causing the problem, or it could be a different problem altogether.

Hope this is helpful!

--The Geek