# Function-like macro

Discussion in 'C Programming' started by Russell Shaw, May 24, 2005.

1. ### Russell ShawGuest

Hi,

How do i make an if/then/else macro act as a function
so that the whole thing looks like the return value?

I tried this lame attempt for starters:

#define A_FROM_B(b) \
( \
if(b < 10) { \
a = b; \
} \
else { \
a = 2*b; \
}, \
a; \
)

Russell Shaw, May 24, 2005

2. ### Richard BosGuest

Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:

> How do i make an if/then/else macro act as a function
> so that the whole thing looks like the return value?

You don't. That's what the ?: operator is for.

> I tried this lame attempt for starters:
>
> #define A_FROM_B(b) \
> ( \
> if(b < 10) { \
> a = b; \

a? What is a? What happens if I call this macro in a function where a is
a struct?

> } \
> else { \
> a = 2*b; \
> }, \
> a; \
> )

#define A_FROM_B(b) ((b)<10? (b): 2*(b))

Note the parens around b in the definition of the macro. They will save
your bacon some day when you decide to call A_FROM_B(x+10).

Richard

Richard Bos, May 24, 2005

3. ### Russell ShawGuest

Richard Bos wrote:
> Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:
>
>>How do i make an if/then/else macro act as a function
>>so that the whole thing looks like the return value?

>
> You don't. That's what the ?: operator is for.
>
>>I tried this lame attempt for starters:
>>
>>#define A_FROM_B(b) \
>> ( \
>> if(b < 10) { \
>> a = b; \

>
> a? What is a? What happens if I call this macro in a function where a is
> a struct?

Like i said, it's a lame one

>> } \
>> else { \
>> a = 2*b; \
>> }, \
>> a; \
>> )

>
> #define A_FROM_B(b) ((b)<10? (b): 2*(b))
>
> Note the parens around b in the definition of the macro. They will save
> your bacon some day when you decide to call A_FROM_B(x+10).
>
> Richard

What do i do with things that contain more complex statements like:

if(size > 0x1ff) {
size_t sz = size;
for(ndx = 0; ndx < 32; ndx++) {
if(sz <= 0) {
break;
}
sz >>= 1;
}
ndx -= 8;
ndx <<= 9;
}
else {
ndx = size;
}

Do i have to make it into a function?

Russell Shaw, May 24, 2005
4. ### Villy KruseGuest

On Tue, 24 May 2005 20:24:28 +1000,
Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:

>
> Do i have to make it into a function?

An inline function wouldn't be such a bad alternative if it is available
with your compiler. Otherwise, "do { ... } while (0)" is a common
ideom for making a function like macro.

Villy

Villy Kruse, May 24, 2005
5. ### -berlin.deGuest

Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:
> Richard Bos wrote:
>> Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:
>>
>>>How do i make an if/then/else macro act as a function
>>>so that the whole thing looks like the return value?

>>
>> You don't. That's what the ?: operator is for.
>>
>>>I tried this lame attempt for starters:
>>>
>>>#define A_FROM_B(b) \
>>> ( \
>>> if(b < 10) { \
>>> a = b; \

>>
>> a? What is a? What happens if I call this macro in a function where a is
>> a struct?

> Like i said, it's a lame one

>>> } \
>>> else { \
>>> a = 2*b; \
>>> }, \
>>> a; \
>>> )

>>
>> #define A_FROM_B(b) ((b)<10? (b): 2*(b))
>>
>> Note the parens around b in the definition of the macro. They will save
>> your bacon some day when you decide to call A_FROM_B(x+10).
>>
>> Richard

> What do i do with things that contain more complex statements like:

> if(size > 0x1ff) {
> size_t sz = size;
> for(ndx = 0; ndx < 32; ndx++) {
> if(sz <= 0) {
> break;
> }
> sz >>= 1;
> }
> ndx -= 8;
> ndx <<= 9;
> }
> else {
> ndx = size;
> }

> Do i have to make it into a function?

You can make a macro out of that, e.g.

#defined STRANGE_MACRO( size, ndx ) \
do { \
if ( ( size ) > 0x1ff) { \
size_t sz = ( size ); \
for ( ndx = 0; ndx < 32; ndx++ ) { \
if ( sz <= 0 ) \
break; \
sz >>= 1; \
} \
ndx -= 8; \
ndx <<= 9; \
} \
else \
ndx = ( size ); \
} while ( 0 ) \

but I would rather recommend to make that a function. Macros can
easily break when not written and used very carefully. If you e.g.
call it as

STRANGE_MACRO( size++, ndx );

then 'size' gets incremented two or three times instead of just
once as you would expect from a function call because a macro i
a simple text replacement (you could avoid that here by introdu-
cing another variable in the block and assigning the value of
'size' to it just once instead of evaluating it several times
in the macro). And, of course, if you try to call it like this

STRANGE_MACRO( size, ndx + 1 );

you'll ge a syntax error that's probably very hard to find. So why
not make the whole thing into a function?

BTW, in that macro you use a shift operation on 'sz' but also test
it for being less or equal to 0. But you can only use shift operations
safely on unsigned integers.
Regards, Jens
--
\ Jens Thoms Toerring ___ -berlin.de
\__________________________ http://www.toerring.de

-berlin.de, May 24, 2005
6. ### CBFalconerGuest

Russell Shaw wrote:
>
> How do i make an if/then/else macro act as a function
> so that the whole thing looks like the return value?
>
> I tried this lame attempt for starters:
>
> #define A_FROM_B(b) \
> ( \
> if(b < 10) { \
> a = b; \
> } \
> else { \
> a = 2*b; \
> }, \
> a; \
> )

try:

#define A_FROM_B(b) do {\
if (b < 10) a = b; \
else { \
a = 2*b; \
} \
} while (0)

The general form is "do { <stuff> } while (0)", and you are free to
write legal code for <stuff>. It does not return a usable value.

--
<http://www.ungerhu.com/jxh/clc.welcome.txt>
<http://www.eskimo.com/~scs/C-faq/top.html>
<http://benpfaff.org/writings/clc/off-topic.html>
<http://anubis.dkuug.dk/jtc1/sc22/wg14/www/docs/n869/> (C99)
<http://www.dinkumware.com/refxc.html> (C-library}
<http://gcc.gnu.org/onlinedocs/> (GNU docs)

CBFalconer, May 24, 2005
7. ### Richard BosGuest

Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:

> Richard Bos wrote:
> > #define A_FROM_B(b) ((b)<10? (b): 2*(b))
> >
> > Note the parens around b in the definition of the macro. They will save
> > your bacon some day when you decide to call A_FROM_B(x+10).

>
> What do i do with things that contain more complex statements like:
>
> if(size > 0x1ff) {
> size_t sz = size;
> for(ndx = 0; ndx < 32; ndx++) {
> if(sz <= 0) {
> break;
> }
> sz >>= 1;
> }
> ndx -= 8;
> ndx <<= 9;
> }
> else {
> ndx = size;
> }
>
> Do i have to make it into a function?

Have to is strong, but in practice that would be the best solution.
Modern compilers are quite good at optimising, so a function this size
would probably be inlined automatically wherever that is more efficient.

Richard

Richard Bos, May 24, 2005
8. ### Richard BosGuest

Villy Kruse <> wrote:

> On Tue, 24 May 2005 20:24:28 +1000,
> Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:
>
> > Do i have to make it into a function?

>
> An inline function wouldn't be such a bad alternative if it is available
> with your compiler. Otherwise, "do { ... } while (0)" is a common
> ideom for making a function like macro.

True, but it has one drawback: you can't return a value from one, so it
acts as a statement block, not as a function call.

Richard

Richard Bos, May 24, 2005
9. ### SM RyanGuest

Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:
# Hi,
#
# How do i make an if/then/else macro act as a function
# so that the whole thing looks like the return value?

You can turn the following statements into expressions. If you mind the precedence,
you can reduce the number of parentheses.

a; b (a,b)
{a} (a)
if (a) b; else c ((a)?(b)c))
expression; (expression)

With declarations and other statements, you're screwed.

Note that "a = b" is an expression. The statement "a = b;" can be converted to
"a = b".

# #define A_FROM_B(b) \
# ( \
# if(b < 10) { \
# a = b; \
# } \
# else { \
# a = 2*b; \
# }, \
# a; \
# )

( \
(b < 10) ? ( \
a = b \
) \
: ( \
a = 2*b \
), \
a \
)

--
SM Ryan http://www.rawbw.com/~wyrmwif/
Elvis was an artist. But that didn't stop him from joining the service
in time of war. That's why he is the king, and you're a shmuck.

SM Ryan, May 24, 2005
10. ### Tim RentschGuest

Russell Shaw <rjshawN_o@s_pam.netspace.net.au> writes:

> How do i make an if/then/else macro act as a function
> so that the whole thing looks like the return value?
>
> I tried this lame attempt for starters:
>
> #define A_FROM_B(b) \
> ( \
> if(b < 10) { \
> a = b; \
> } \
> else { \
> a = 2*b; \
> }, \
> a; \
> )

Many macros written in statement form can be written fairly easily in
expression form instead. Is there any reason

#define A_FROM_B(b) (a = (b)<10 ? (b) : 2*(b))

doesn't meet your needs in this case?

As a matter of good form I'd usually prefer that the assigned-to

#define ASSIGN_BAROQUELY(a,b) ((a) = (b)<10 ? (b) : 2*(b))

but that is a separate discussion.

Those macro definitions that can't be written in expression form and
that need to return a value are probably better written as functions,
whether inline functions are available or not.

Tim Rentsch, May 24, 2005
11. ### MalcolmGuest

"Russell Shaw" <rjshawN_o@s_pam.netspace.net.au> wrote
> How do i make an if/then/else macro act as a function
> so that the whole thing looks like the return value?
>
> I tried this lame attempt for starters:
>
> #define A_FROM_B(b) \
> ( \
> if(b < 10) { \
> a = b; \
> } \
> else { \
> a = 2*b; \
> }, \
> a; \
> )

#define A_FROM_B(b) ((b) < 10 ? (b) : (2 * (b)))

Malcolm, May 24, 2005
12. ### Dale HagglundGuest

Russell> How do i make an if/then/else macro act as a function so
Russell> that the whole thing looks like the return value?

Others have already pointed out that you're probably much better off
using an inline function. It's more easily readable and editable.
That said, GCC does have a non-standard extension that allows you to
do what you want:

#define A_FROM_B(b) \
({ \
int __b = (b); \
int __a; \
if (__b < 10) \
__a = __b; \
else \
__a = 2*__b; \
__a; \
})

A few notes:

* I haven't compiled the above code so I may have a couple of
typos in it. See Statement Exprs in the gcc info pages.

* I introduced the __b variable to prevent multiple evaluation

* The value of the macro is the the last "bare" expression in
the block.

* As I mentioned, as far as I know this is non-standard.

* GCC has a couple of other nifty extensions that work well
together with this extension. Especially see the gcc info
on typeof().

* Because macros are based on textual substitution, it's a
good idea to introduce local variable names that aren't
likely to collide with names used inside the macro
arguments. (Again, functions completely avoid this issue.)

Dale.

Dale Hagglund, May 25, 2005
13. ### Russell ShawGuest

Dale Hagglund wrote:
> Russell> How do i make an if/then/else macro act as a function so
> Russell> that the whole thing looks like the return value?
>
> Others have already pointed out that you're probably much better off
> using an inline function. It's more easily readable and editable.
> That said, GCC does have a non-standard extension that allows you to
> do what you want:
>
> #define A_FROM_B(b) \
> ({ \
> int __b = (b); \
> int __a; \
> if (__b < 10) \
> __a = __b; \
> else \
> __a = 2*__b; \
> __a; \
> })
>
> A few notes:
>
> * I haven't compiled the above code so I may have a couple of
> typos in it. See Statement Exprs in the gcc info pages.
>
> * I introduced the __b variable to prevent multiple evaluation
>
> * The value of the macro is the the last "bare" expression in
> the block.
>
> * As I mentioned, as far as I know this is non-standard.
>
> * GCC has a couple of other nifty extensions that work well
> together with this extension. Especially see the gcc info
> on typeof().
>
> * Because macros are based on textual substitution, it's a
> good idea to introduce local variable names that aren't
> likely to collide with names used inside the macro
> arguments. (Again, functions completely avoid this issue.)
>
> Dale.

Interesting, it is in gcc-3.4 info help 5.1 Statements and Declarations in Expressions.

Does a curly-bracket block {...} have a value in standard C? I've been looking for
a long time.

The macro above works (i tested it).

Russell Shaw, May 25, 2005
14. ### Dale HagglundGuest

[My example of gcc statement expression syntax deleted. --rdh]

Russell> Does a curly-bracket block {...} have a value in standard
Russell> C?

Nope. As far as I know its a gcc-only extension.

Dale.

Dale Hagglund, May 25, 2005
15. ### -berlin.deGuest

Russell Shaw <rjshawN_o@s_pam.netspace.net.au> wrote:
> Dale Hagglund wrote:
>> * The value of the macro is the the last "bare" expression in
>> the block.
>>
>> * As I mentioned, as far as I know this is non-standard.

> Interesting, it is in gcc-3.4 info help 5.1 Statements and Declarations in Expressions.

> Does a curly-bracket block {...} have a value in standard C? I've been
> looking for a long time.

No, a block (compound statement) isn't an expression and thus has no
value. So something like

int i = { 1; };

or

int i = ( { 1; } );

or variations of these aren't standard C. gcc lets you use the second
version, i.e. a compound statement enclosed in parentheses where the
value of the last expression in the block is taken to be the value of
of the whole construct, as an extension only (that's why it is listed
in the info pages under "C Extensions";-)

Regards, Jens
--
\ Jens Thoms Toerring ___ -berlin.de
\__________________________ http://www.toerring.de

-berlin.de, May 25, 2005
16. ### Richard BosGuest

Dale Hagglund <> wrote:

> Others have already pointed out that you're probably much better off
> using an inline function. It's more easily readable and editable.
> That said, GCC does have a non-standard extension that allows you to
> do what you want:
>
> #define A_FROM_B(b) \
> ({ \
> int __b = (b); \
> int __a; \
> if (__b < 10) \
> __a = __b; \
> else \
> __a = 2*__b; \
> __a; \
> })

> * I introduced the __b variable to prevent multiple evaluation

Unfortunately it also makes the macro non-portable, since all
identifiers starting with __ (or _ plus capital letter) are reserved for
the implementation. __b is simple enough that it could clash with
something defined in a system header.

> * As I mentioned, as far as I know this is non-standard.

'tis.

> * Because macros are based on textual substitution, it's a
> good idea to introduce local variable names that aren't
> likely to collide with names used inside the macro
> arguments. (Again, functions completely avoid this issue.)

But make sure they're yours to use.

Richard

Richard Bos, May 26, 2005
17. ### peteGuest

Richard Bos wrote:

> #define A_FROM_B(b) ((b)<10? (b): 2*(b))
>
> Note the parens around b in the definition of the macro.
> They will save
> your bacon some day when you decide to call A_FROM_B(x+10).

Trivia point:
Only the first and last pair are needed for bacon saving.

#define A_FROM_B(b) ((b) < 10 ? b : 2 * (b))

? b :
means the same thing as
?(b):

--
pete

pete, May 27, 2005
18. ### Richard BosGuest

pete <> wrote:

> Richard Bos wrote:
>
> > #define A_FROM_B(b) ((b)<10? (b): 2*(b))
> >
> > Note the parens around b in the definition of the macro.
> > They will save
> > your bacon some day when you decide to call A_FROM_B(x+10).

>
> Trivia point:
> Only the first and last pair are needed for bacon saving.
>
> #define A_FROM_B(b) ((b) < 10 ? b : 2 * (b))
>
> ? b :
> means the same thing as
> ?(b):

In this case, perhaps. Howsoever, I'd advise getting into the habit of
parenthesisising anyway, for psychobabblical reasons.

Richard

Richard Bos, May 27, 2005