How do I make a C++ macro behave like a function? - c++

Question

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. :)

Answer 1

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.

Answer 2

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.

Answer 3

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 warranted.
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.

Answer 4

Here is an answer coming right from the libc6!
Taking a look at /usr/include/x86_64-linux-gnu/bits/byteswap.h, I found the trick you were looking for.
A few critics of previous solutions:
Kip's solution does not permit evaluating to an expression, which is in the end often needed.
coppro's solution does not permit assigning a variable as the expressions are separate, but can evaluate to an expression.
Steve Jessop's solution uses the C++11 auto keyword, that's fine, but feel free to use the known/expected type instead.
The trick is to use both the (expr,expr) construct and a {} scope:
#define MACRO(X,Y) \
( \
{ \
register int __x = static_cast<int>(X), __y = static_cast<int>(Y); \
std::cout << "1st arg is:" << __x << std::endl; \
std::cout << "2nd arg is:" << __y << std::endl; \
std::cout << "Sum is:" << (__x + __y) << std::endl; \
__x + __y; \
} \
)
Note the use of the register keyword, it's only a hint to the compiler.
The X and Y macro parameters are (already) surrounded in parenthesis and casted to an expected type.
This solution works properly with pre- and post-increment as parameters are evaluated only once.
For the example purpose, even though not requested, I added the __x + __y; statement, which is the way to make the whole bloc to be evaluated as that precise expression.
It's safer to use void(); if you want to make sure the macro won't evaluate to an expression, thus being illegal where an rvalue is expected.
However, the solution is not ISO C++ compliant as will complain g++ -pedantic:
warning: ISO C++ forbids braced-groups within expressions [-pedantic]
In order to give some rest to g++, use (__extension__ OLD_WHOLE_MACRO_CONTENT_HERE) so that the new definition reads:
#define MACRO(X,Y) \
(__extension__ ( \
{ \
register int __x = static_cast<int>(X), __y = static_cast<int>(Y); \
std::cout << "1st arg is:" << __x << std::endl; \
std::cout << "2nd arg is:" << __y << std::endl; \
std::cout << "Sum is:" << (__x + __y) << std::endl; \
__x + __y; \
} \
))
In order to improve my solution even a bit more, let's use the __typeof__ keyword, as seen in MIN and MAX in C:
#define MACRO(X,Y) \
(__extension__ ( \
{ \
__typeof__(X) __x = (X); \
__typeof__(Y) __y = (Y); \
std::cout << "1st arg is:" << __x << std::endl; \
std::cout << "2nd arg is:" << __y << std::endl; \
std::cout << "Sum is:" << (__x + __y) << std::endl; \
__x + __y; \
} \
))
Now the compiler will determine the appropriate type. This too is a gcc extension.
Note the removal of the register keyword, as it would the following warning when used with a class type:
warning: address requested for ‘__x’, which is declared ‘register’ [-Wextra]

Answer 5

C++11 brought us lambdas, which can be incredibly useful in this situation:
#define MACRO(X,Y) \
[&](x_, y_) { \
cout << "1st arg is:" << x_ << endl; \
cout << "2nd arg is:" << y_ << endl; \
cout << "Sum is:" << (x_ + y_) << endl; \
}((X), (Y))
You keep the generative power of macros, but have a comfy scope from which you can return whatever you want (including void). Additionally, the issue of evaluating macro parameters multiple times is avoided.

Answer 6

Create a block using
#define MACRO(...) do { ... } while(false)
Do not add a ; after the while(false)

Answer 7

Your answer suffers from the multiple-evaluation problem, so (eg)
macro( read_int(file1), read_int(file2) );
will do something unexpected and probably unwanted.

Answer 8

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

Answer 9

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();