Weird bit of code-- legal???

[Ancient_Hacker]

A while back I had to recompile some old code, originally written by a
really good programmer, but one prone to use every trick in the book,
and then some.

There was a statement, something like this:
Result = ( Mode == 2 ? TwoWayFunc : FourWayFunc ) ( Param1,
Param2, Param3 );

... after a little headscratching I figured it out, but ... is this
legal in C? It compiled with an Microsoft C 2.6 or so compiler, but
of course that isnt much help.

[gbostock]

Ancient_Hacker wrote:

A while back I had to recompile some old code, originally written by a
really good programmer, but one prone to use every trick in the book,
and then some.

There was a statement, something like this:
Result = ( Mode == 2 ? TwoWayFunc : FourWayFunc ) ( Param1,
Param2, Param3 );

... after a little headscratching I figured it out, but ... is this
legal in C? It compiled with an Microsoft C 2.6 or so compiler, but
of course that isnt much help.

This code is really quite simple and shouldn't give a decent C
programmer any problems whatsoever.

The line is basically a function call that puts the result in Result.
Depending on the value of Mode (is it 2?) it will invoke the function
pointed to by TwoWayFunc or FourWayFunc; passing the Params 1 through
3.

Why shouldn't it be legal and why would it cause any head scratching?

[Ancient_Hacker]

gbost...@excite .com wrote:

Ancient_Hacker wrote:

A while back I had to recompile some old code, originally written by a
really good programmer, but one prone to use every trick in the book,
and then some.

There was a statement, something like this:
Result = ( Mode == 2 ? TwoWayFunc : FourWayFunc ) ( Param1,
Param2, Param3 );

... after a little headscratching I figured it out, but ... is this
legal in C? It compiled with an Microsoft C 2.6 or so compiler, but
of course that isnt much help.

This code is really quite simple and shouldn't give a decent C
programmer any problems whatsoever.

The line is basically a function call that puts the result in Result.
Depending on the value of Mode (is it 2?) it will invoke the function
pointed to by TwoWayFunc or FourWayFunc; passing the Params 1 through
3.

Why shouldn't it be legal and why would it cause any head scratching?

.... because I didnt mention, the function names are just that-- not
variables holding the function addresses. I did not know you could
use raw function names in a conditional expression. Then again, I
didnt know and I suspect a lot of folks don't know that the thing
before the parameters is an actual "expression ".

[Frederick Gotham]

Ancient_Hacker posted:

Result = ( Mode == 2 ? TwoWayFunc : FourWayFunc ) ( Param1,
Param2, Param3 );

... after a little headscratching I figured it out, but ... is this
legal in C?

Not to boast, but I understood the line of code at first glance.

"TwoWayFunc " and "FourWayFun c" are pressumably both function pointers of the
same type.

"Mode == 2" is evaluated, and depending on the result, either of the
functions is called with the arguments, "Param1, Param2, Param3".

The return value of the function call is stored in "Result".

The code is perfectly legal.

--

Frederick Gotham

[Frederick Gotham]

Ancient_Hacker posted:

... because I didnt mention, the function names are just that-- not
variables holding the function addresses. I did not know you could
use raw function names in a conditional expression.

When you use the name of a function on its own (i.e. without parenthese after
it which contain arguments), then what you have is the address of that
function.

For instance, let's define a function:

void Func1(void) {}

Now let's create a pointer to a function of that type:

void (*p)(void);

Now let's store the address of Func1 in p:

p = Func1;

The expression "Func1" in the above is the address of the function.

--

Frederick Gotham

[Kenny McCormack]

In article <11************ **********@p79g 2000cwp.googleg roups.com>,
<gb******@excit e.comwrote:
....

>There was a statement, something like this:
Result = ( Mode == 2 ? TwoWayFunc : FourWayFunc ) ( Param1,
Param2, Param3 );

... after a little headscratching I figured it out, but ... is this
legal in C? It compiled with an Microsoft C 2.6 or so compiler, but
of course that isnt much help.

This code is really quite simple and shouldn't give a decent C
programmer any problems whatsoever.

It is not a question of "understand ing" it or whether or not it works on
any given implementation. As you well know, both of those topics are OT
in this ng.

Rather, it is a question of whether or not it is "strictly conforming",
according to the gods of the ng. I remember reading here sometime back
that there are some funky rules regarding "?:" expressions in the
standard - of the usual sort, that is, that allow to fail things that
obviously should work and which do work on most-if-not-all implementations .

And which are thus, by the rules of the NG,
OT/non-portable/wrong/can't-discuss-here/blah-blah-blah.

[Ancient_Hacker]

Frederick Gotham wrote:

Not to boast, but I understood the line of code at first glance.

"TwoWayFunc " and "FourWayFun c" are pressumably both function pointers of the
same type.

That's the funny thing, they're not ptrs. They're the raw function
names. I'd never seen conditional expressions used in that particular
context.

[jmcgill]

Ancient_Hacker wrote:

Frederick Gotham wrote:

>Not to boast, but I understood the line of code at first glance.

"TwoWayFunc " and "FourWayFun c" are pressumably both function pointers of the
same type.

That's the funny thing, they're not ptrs.

Doesn't that make them constant pointers? Or are they different?

Thumbing through K&R2 to try to answer this myself.

[Richard Heathfield]

Ancient_Hacker said:

>
Frederick Gotham wrote:

>Not to boast, but I understood the line of code at first glance.

"TwoWayFunc " and "FourWayFun c" are pressumably both function pointers of
the same type.

That's the funny thing, they're not ptrs.

Yes, they are.

They're the raw function names.

In other words, they're pointers.

Look up "function designator" in the Standard.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

[Nelu]

Ancient_Hacker wrote:

Frederick Gotham wrote:

>Not to boast, but I understood the line of code at first glance.

"TwoWayFunc " and "FourWayFun c" are pressumably both function pointers of the
same type.

That's the funny thing, they're not ptrs. They're the raw function
names. I'd never seen conditional expressions used in that particular
context.

C99, 6.3.2 Other operands
4. A function designator is an expression that has function type.
Except when it is the
operand of the sizeof operator or the unary & operator, a
function designator with
type ‘‘function returning type’’ is converted to an expression
that has type ‘‘pointer to
function returning type’’.

--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)