In many C/C++ macros I\'m seeing the code of the macro wrapped in what seems like a meaningless do while
loop. Here are examples.
#define FOO(X
Macros are copy/pasted pieces of text the pre-processor will put in the genuine code; the macro's author hopes the replacement will produce valid code.
There are three good "tips" to succeed in that:
Normal code is usually ended by a semi-colon. Should the user view code not needing one...
doSomething(1) ;
DO_SOMETHING_ELSE(2) // <== Hey? What's this?
doSomethingElseAgain(3) ;
This means the user expects the compiler to produce an error if the semi-colon is absent.
But the real real good reason is that at some time, the macro's author will perhaps need to replace the macro with a genuine function (perhaps inlined). So the macro should really behave like one.
So we should have a macro needing semi-colon.
As shown in jfm3's answer, sometimes the macro contains more than one instruction. And if the macro is used inside a if statement, this will be problematic:
if(bIsOk)
MY_MACRO(42) ;
This macro could be expanded as:
#define MY_MACRO(x) f(x) ; g(x)
if(bIsOk)
f(42) ; g(42) ; // was MY_MACRO(42) ;
The g
function will be executed regardless of the value of bIsOk
.
This means that we must have to add a scope to the macro:
#define MY_MACRO(x) { f(x) ; g(x) ; }
if(bIsOk)
{ f(42) ; g(42) ; } ; // was MY_MACRO(42) ;
If the macro is something like:
#define MY_MACRO(x) int i = x + 1 ; f(i) ;
We could have another problem in the following code:
void doSomething()
{
int i = 25 ;
MY_MACRO(32) ;
}
Because it would expand as:
void doSomething()
{
int i = 25 ;
int i = 32 + 1 ; f(i) ; ; // was MY_MACRO(32) ;
}
This code won't compile, of course. So, again, the solution is using a scope:
#define MY_MACRO(x) { int i = x + 1 ; f(i) ; }
void doSomething()
{
int i = 25 ;
{ int i = 32 + 1 ; f(i) ; } ; // was MY_MACRO(32) ;
}
The code behaves correctly again.
There is one C/C++ idiom that produces this effect: The do/while loop:
do
{
// code
}
while(false) ;
The do/while can create a scope, thus encapsulating the macro's code, and needs a semi-colon in the end, thus expanding into code needing one.
The bonus?
The C++ compiler will optimize away the do/while loop, as the fact its post-condition is false is known at compile time. This means that a macro like:
#define MY_MACRO(x) \
do \
{ \
const int i = x + 1 ; \
f(i) ; g(i) ; \
} \
while(false)
void doSomething(bool bIsOk)
{
int i = 25 ;
if(bIsOk)
MY_MACRO(42) ;
// Etc.
}
will expand correctly as
void doSomething(bool bIsOk)
{
int i = 25 ;
if(bIsOk)
do
{
const int i = 42 + 1 ; // was MY_MACRO(42) ;
f(i) ; g(i) ;
}
while(false) ;
// Etc.
}
and is then compiled and optimized away as
void doSomething(bool bIsOk)
{
int i = 25 ;
if(bIsOk)
{
f(43) ; g(43) ;
}
// Etc.
}
For some reasons I can't comment on the first answer...
Some of you showed macros with local variables, but nobody mentioned that you can't just use any name in a macro! It will bite the user some day! Why? Because the input arguments are substituted into your macro template. And in your macro examples you've use the probably most commonly used variabled name i.
For example when the following macro
#define FOO(X) do { int i; for (i = 0; i < (X); ++i) do_something(i); } while (0)
is used in the following function
void some_func(void) {
int i;
for (i = 0; i < 10; ++i)
FOO(i);
}
the macro will not use the intended variable i, that is declared at the beginning of some_func, but the local variable, that is declared in the do ... while loop of the macro.
Thus, never use common variable names in a macro!
The do ... while
and if ... else
are there to make it so that a
semicolon after your macro always means the same thing. Let's say you
had something like your second macro.
#define BAR(X) f(x); g(x)
Now if you were to use BAR(X);
in an if ... else
statement, where the bodies of the if statement were not wrapped in curly brackets, you'd get a bad surprise.
if (corge)
BAR(corge);
else
gralt();
The above code would expand into
if (corge)
f(corge); g(corge);
else
gralt();
which is syntactically incorrect, as the else is no longer associated with the if. It doesn't help to wrap things in curly braces within the macro, because a semicolon after the braces is syntactically incorrect.
if (corge)
{f(corge); g(corge);};
else
gralt();
There are two ways of fixing the problem. The first is to use a comma to sequence statements within the macro without robbing it of its ability to act like an expression.
#define BAR(X) f(X), g(X)
The above version of bar BAR
expands the above code into what follows, which is syntactically correct.
if (corge)
f(corge), g(corge);
else
gralt();
This doesn't work if instead of f(X)
you have a more complicated body of code that needs to go in its own block, say for example to declare local variables. In the most general case the solution is to use something like do ... while
to cause the macro to be a single statement that takes a semicolon without confusion.
#define BAR(X) do { \
int i = f(X); \
if (i > 4) g(i); \
} while (0)
You don't have to use do ... while
, you could cook up something with if ... else
as well, although when if ... else
expands inside of an if ... else
it leads to a "dangling else", which could make an existing dangling else problem even harder to find, as in the following code.
if (corge)
if (1) { f(corge); g(corge); } else;
else
gralt();
The point is to use up the semicolon in contexts where a dangling semicolon is erroneous. Of course, it could (and probably should) be argued at this point that it would be better to declare BAR
as an actual function, not a macro.
In summary, the do ... while
is there to work around the shortcomings of the C preprocessor. When those C style guides tell you to lay off the C preprocessor, this is the kind of thing they're worried about.