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8

Let's say that for some reason you need to write a macro: MACRO(X,Y). (Let's assume there's a good reason you can't use an inline function.) You want this macro to emulate a call to a function with no return value.


Example 1: This should work as expected.

if (x > y)
  MACRO(x, y);
do_something();

Example 2: This should not result in a compiler error.

if (x > y)
  MACRO(x, y);
else
  MACRO(y - x, x - y);

Example 3: This should not compile.

do_something();
MACRO(x, y)
do_something();


The naïve way to write the macro is like this:

#define MACRO(X,Y)                       \
cout << "1st arg is:" << (X) << endl;    \
cout << "2nd arg is:" << (Y) << endl;    \
cout << "Sum is:" << ((X)+(Y)) << endl;

This is a very bad solution which fails all three examples, and I shouldn't need to explain why.

Ignore what the macro actually does, that's not the point.


Now, the way I most often see macros written is to enclose them in curly braces, like this:

#define MACRO(X,Y)                         \
{                                          \
  cout << "1st arg is:" << (X) << endl;    \
  cout << "2nd arg is:" << (Y) << endl;    \
  cout << "Sum is:" << ((X)+(Y)) << endl;  \
}

This solves example 1, because the macro is in one statement block. But example 2 is broken because we put a semicolon after the call to the macro. This makes the compiler think the semicolon is a statement by itself, which means the else statement doesn't correspond to any if statement! And lastly, example 3 compiles OK, even though there is no semicolon, because a code block doesn't need a semicolon.


Is there a way to write a macro so that it pass all three examples?


Note: I am submitting my own answer as part of the accepted way of sharing a tip, but if anyone has a better solution feel free to post it here, it may get more votes than my method. :)

+12  A: 

There is a rather clever solution:

#define MACRO(X,Y)                         \
do {                                       \
  cout << "1st arg is:" << (X) << endl;    \
  cout << "2nd arg is:" << (Y) << endl;    \
  cout << "Sum is:" << ((X)+(Y)) << endl;  \
} while (0)

Now you have a single block-level statement, which must be followed by a semicolon. This behaves as expected and desired in all three examples.

Kip
How about "MACRO(++x, --y)"?
Fred Larson
Yes we all know the pitfalls of calling a macro with complicated arguments. That's why everyone does (or should!) name macros in all caps, so you know not to do anything stupid.
Kip
The multiple evaluation problem isn't insurmountable, btw. Check this code a smart guy did --> http://pastie.org/357910 It provides a foolproof "RANGE(foo)" macro that generates "foo.begin(), foo.end()" basically, but it's applicable for addressing any multiple evaluation. Runtime overhead, though.
+1  A: 

Create a block using

 #define MACRO(...) do { ... } while(false)

Do not add a ; after the while(false)

Andrew Stein
A: 

If you're willing to adopt the practice of always using curly braces in your if statements,

Your macro would simply be missing the last semicolon:

#define MACRO(X,Y)                       \
cout << "1st arg is:" << (X) << endl;    \
cout << "2nd arg is:" << (Y) << endl;    \
cout << "Sum is:" << ((X)+(Y)) << endl

Example 1: (compiles)

if (x > y) {
    MACRO(x, y);
}
do_something();

Example 2: (compiles)

if (x > y) {
    MACRO(x, y);
} else {
    MACRO(y - x, x - y);
}

Example 3: (doesn't compile)

do_something();
MACRO(x, y)
do_something();
Ates Goral
+10  A: 

Macros should generally be avoided; prefer inline functions to them at all times. Any compiler worth its salt should be capable of inlining a small function as if it were a macro, and an inline function will respect namespaces and other scopes, as well as evaluating all the arguments once.

If it must be a macro, a while loop (already suggested) will work, or you can try the comma operator:

#define MACRO(X,Y) \
 ( \
  (cout << "1st arg is:" << (X) << endl), \
  (cout << "2nd arg is:" << (Y) << endl), \
  (cout << "3rd arg is:" << ((X) + (Y)) << endl), \
  (void)0 \
 )

The (void)0 causes the statement to evaluate to one of void type, and the use of commas rather than semicolons allows it to be used inside a statement, rather than only as a standalone. I would still recommend an inline function for a host of reasons, the least of which being scope and the fact that MACRO(a++, b++) will increment a and b twice.

coppro
Excellent, this is even better than what I had originally done! Thanks!
Kip
Functions are also superior in that the compiler can decide whether to inline them or not, whereas it has no choice with macros. Inlining is very often not the performance win that people expect.
"Excellent, this is even better than what I had originally done!" -- You're wrong Kip, your answer is better than this one. Yours can behave like more kinds of functions than this one, for example if it needs a loop in the middle of it.
Windows programmer
Kip, your version is better. If you have to declare some local variables in your macro, declaring them inside a while block won't make them leak into the calling scope.
Vincent Robert
+1  A: 

Your answer suffers from the multiple-evaluation problem, so (eg)

macro( read_int(file1), read_int(file2) );

will do something unexpected and probably unwanted.

A: 

As others have mentioned, you should avoid macros whenever possible. They are dangerous in the presence of side effects if the macro arguments are evaluated more than once. If you know the type of the arguments (or can use C++0x auto feature), you could use temporaries to enforce single evaluation.

Another problem: the order in which multiple evaluations happen may not be what you expect!

Consider this code:

#include <iostream>
using namespace std;

int foo( int & i ) { return i *= 10; }
int bar( int & i ) { return i *= 100; }

#define BADMACRO( X, Y ) do { \
    cout << "X=" << (X) << ", Y=" << (Y) << ", X+Y=" << ((X)+(Y)) << endl; \
    } while (0)

#define MACRO( X, Y ) do { \
    int x = X; int y = Y; \
    cout << "X=" << x << ", Y=" << y << ", X+Y=" << ( x + y ) << endl; \
    } while (0)

int main() {
    int a = 1; int b = 1;
    BADMACRO( foo(a), bar(b) );
    a = 1; b = 1;
    MACRO( foo(a), bar(b) );
    return 0;
}

And it's output as compiled and run on my machine:

X=100, Y=10000, X+Y=110
X=10, Y=100, X+Y=110
jwfearn
The problem with your good macro is that you're assuming X and Y are int's. They could be anything. You'd need to templatize your macro or something.
Tanktalus
+4  A: 

I know you said "ignore what the macro does", but people will find this question by searching based on the title, so I think discussion of further techniques to emulate functions with macros are warrented.

Closest I know of is:

#define MACRO(X,Y) \
do { \
    auto MACRO_tmp_1 = (X); \
    auto MACRO_tmp_2 = (Y); \
    using std::cout; \
    using std::endl; \
    cout << "1st arg is:" << (MACRO_tmp_1) << endl;    \
    cout << "2nd arg is:" << (MACRO_tmp_2) << endl;    \
    cout << "Sum is:" << (MACRO_tmp_1 + MACRO_tmp_2) << endl; \
} while(0)

This does the following:

  • Works correctly in each of the stated contexts.
  • Evaluates each of its arguments exactly once, which is a guaranteed feature of a function call (assuming in both cases no exceptions in any of those expressions).
  • Acts on any types, by use of "auto" from C++0x. This is not yet standard C++, but there's no other way to get the tmp variables necessitated by the single-evaluation rule.
  • Doesn't require the caller to have imported names from namespace std, which the original macro does, but a function would not.

However, it still differs from a function in that:

  • In some invalid uses it may give different compiler errors or warnings.
  • It goes wrong if X or Y contain uses of 'MACRO_tmp_1' or 'MACRO_tmp_2' from the surrounding scope.
  • Related to the namespace std thing: a function uses its own lexical context to look up names, whereas a macro uses the context of its call site. There's no way to write a macro that behaves like a function in this respect.
  • It can't be used as the return expression of a void function, which a void expression (such as the comma solution) can. This is even more of an issue when the desired return type is not void, especially when used as an lvalue. But the comma solution can't include using declarations, because they're statements, so pick one or use the ({ ... }) GNU extension.
Steve Jessop