do..while(0) in macro of LINUX kernel

I’ve found link to this interesting article from traxacun’s blog. I am very astonished about this code technique. Wow.. Kernel hackers are prodigies.

Why don’t you ready this article right now?



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

There are a couple of reasons:

* (from Dave Miller) Empty statements give a warning from the compiler so this is why you see #define FOO do { } while(0).
* (from Dave Miller) It gives you a basic block in which to declare local variables.
* (from Ben Collins) 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); \

Now imagine using it like:

if (blah == 2)

This interprets to:

if (blah == 2)
printf(“arg is %s\n”, 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);
} while (0);

Which is exactly what you want.
* (from Per Persson) As both Miller and Collins point out, you want a block statement so you can have several lines of code and declare local variables. But then the natural thing would be to just use for 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 if-statement with two branches:

exch(x,y); // Branch 1
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”

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:

do {
int tmp;
tmp = x;
x = y;
y = tmp;
} while(0);

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