Why do a lot of #defines in the kernel use do { ... } while(0)?

Question: Why do a lot of #defines in the kernel use do { ... } while(0)?

Answer:

There are a couple of reasons:

• Empty statements give a warning from the compiler so this is why

you see #define FOO do { } while(0).

• It gives you a basic block in which to declare local variables.

• It allows you to use more complex macros in conditional code.

Imagine a macro of several lines of code like:

#define FOO(x) \

printf("arg is %s\n", x); \

do_something_useful(x);

Now imagine using it like:

if (blah == 2)

FOO(blah);

This interprets to:

if (blah == 2)

printf("arg is %s\n", blah);

do_something_useful(blah);;

As you can see, the if then only encompasses the printf(), and the do_something_useful()

call is unconditional (not within the scope of the if), like you wanted it. So, by using a

block like do { ... } while(0), you would get this:

if (blah == 2)

do {

printf("arg is %s\n", blah);

do_something_useful(blah);

} while (0);

Which is exactly what you want.

•Another example:

#define exch(x,y) { int tmp; tmp=x; x=y; y=tmp; }

However that wouldn't work in some cases. The following code is meant to be an ifstatement

with two branches:

if (x > y)

exch(x,y); // Branch 1

else

do_something(); // Branch 2

But it would be interpreted as an if-statement with only one branch:

if (x > y) { // Single-branch if-statement!!!

int tmp; // The one and only branch consists

tmp = x; // of the block.

x = y;

y = tmp;

}

; // empty statement

else // ERROR!!! "parse error before else"

do_something();

The problem is the semi-colon (;) coming directly after the block. The solution for this

is to sandwich the block between do and while (0). Then we have a single

statement with the capabilities of a block, but not considered as being a block statement

by the compiler. Our if-statement now becomes:

if (x > y)

do {

int tmp;

tmp = x;

x = y;

y = tmp;

} while(0);

else

do_something();

Bye...

Umakanta