Yan wrote:
A lot of times when reading open software, i come across macros that are
defined as follows:
#define CALL_FUNCS(x) \
do { \
func1(x); \
func2(x); \
func3(x); \
} while (0);
now, of course this will work but how is this any better than:
#define CALL_FUNCS(x) \
{ \
func1(x); \
func2(x); \
func3(x); \
}
i can't see how the compiler can optimize (a) any better than (b) or in
any case can (b) break what (a) won't. Any input will be appreciated.
It is not about optimization.
The whole idea of using 'do/while' version is to make a macro which will
expand into a regular statement, not into a compound statement. This is
done in order to make the use of function-style macros uniform with the
use of ordinary functions in all contexts.
Consider the following code sketch
if (<condition>)
foo(a);
else
bar(a);
where 'foo' and 'bar' are ordinary functions. Now imagine that you'd
like to replace function 'foo' with a macro of the above nature
if (<condition>)
CALL_FUNCS(a);
else
bar(a);
Now, if your macro is defined in accordance with the second approach
(just '{' and '}') the code will no longer compile, because the 'true'
branch of 'i' is now represented by a compound statement. And when you
put a ';' after this compound statement, you finished the whole 'if'
statement, thus orphaning the 'else' branch (hence the compilation error).
One way to correct this problem is to remember not to put ';' after
macro "invocation s"
if (<condition>)
CALL_FUNCS(a)
else
bar(a);
This will compile and work as expected, but this is not uniform. The
more elegant solution is to make sure that macro expand into a regular
statement, not into a compound one. One way to achieve that is to define
the macro as follows
#define CALL_FUNCS(x) \
do { \
func1(x); \
func2(x); \
func3(x); \
} while (0)
Now this code
if (<condition>)
CALL_FUNCS(a);
else
bar(a);
will compile without any problems.
However, note the small but important difference between my definition
of 'CALL_FUNCS' and the first version in your message. I didn't put a
';' after '} while (0)'. Putting a ';' at the end of that definition
would immediately defeat the entire point of using 'do/while' and make
that macro pretty much equivalent to the compound-statement version.
I don't know why the author of the code you quoted in your original
message put this ';' after 'while (0)'. In this form both variants are
equivalent. The whole idea behind using 'do/while' version is not to
include this final ';' into the macro (for the reasons that I explained
above).
--
Best regards,
Andrey Tarasevich
