McEs, A Hacker Life: Pass by Reference in C

Monday, August 08, 2005

Pass by Reference in C

When reading setjmp.h I figured out a nice trick used by glibc hackers to implement pass-by-reference'd data types. It was like Wow! Cool! Beautiful! Automatic memory allocation with no reference operator when passing to functions. This has been indeed part of the design of C and exactly why array types were added to C, yet I never figured out arrays can be used to implement opaque types that are passed by reference.

#include <stdio.h>

typedef struct __dummy_tag {
 int x;
} the_type[1];


void
setter(the_type i, int v)
{
 i->x = v;
}

int
getter(the_type i)
{
 return i->x;
}


int
main(void)
{
 the_type v;

 setter (v, 10);
 printf ("%d\n", getter (v));

 setter (v, 20);
 printf ("%d\n", getter (v));

 return 0;
}

You learn new things everyday...

Update: Apparently blogger has !@#$ed up my post. Kindly read the code syntax-highlighted and cutely-indented in my blog.

Update2: Originally I had planet blamed instead of blogger in the previous update. Giving up after a few tries to do overstrike on planet, I simply replaced it with blogger.

¶ 2:59 AM

Comments:

behdad, Planet did precisely what your atom feed told it to. :-)

# posted by

Jeff Waugh : August 08, 2005 4:32 AM

I'd say it's not not much "pass-by-reference" -- you can achieve the same effect by using pointers for function parameters. The trick just allows to avoid '&' operator everywhere, and so can be used not only for parameter passing. Say, I often use this:

void string_new( string* );
void string_free( string* );
......
string s[1];
string_new(s);
........
string_free(s);

just as syntactic convenience.

# posted by

Vladimir Prus : August 08, 2005 6:13 AM

But doing that way, you get convenience only when passing that variable as function parameter, while doing a direct assignment will require extra syntax. I think that it's a nice trick only if used with structs.

# posted by

Anonymous : August 08, 2005 8:21 AM

A slightly more readable version of the typedef would be:

typedef struct __dummy_tag {
int x;
} * the_type;

BTW, I don't like this pass-by-ref-without-ampersands trick because it obscures the intention of the code. It's not immediately obvious (other than by its name) that setter will alter the contents of v -- this can very confusing for a maintainer (not the original author of the code) who is for sure accustomed to see '&' for pass-by-reference in C and could easily overlook the side-effects of 'setter'

# posted by

Anonymous : August 08, 2005 11:53 AM

Starting from the first comment so far:

Jeff: Sorry man, fixed the notice.

Vladimir: The whole pass-by-reference thing /is/ syntactic sugar. No wonder we don't have it in C, because it's only syntactically different from passing a pointer.

Anonymous: Right. You are fine as long as you treat it as an opaque type and don't assign it. Otherwise you need to treat it like a pointer.

Next Anonymous: Man, the whole trick is to use an array of length 1 instead of a pointer. If you typedef to a pointer, then defining a variable of that type does not allocate memory for it, you have to malloc(). About obscuring, true, it breaks all hidden assumptions of a C programmer, but then it's just starting to look like C++...

# posted by

behdad : August 08, 2005 3:22 PM

Why

typedef struct __dummy_tag {
int x;
} the_type[1];

and not

typedef struct __dummy_tag {
int x;
} the_type[0];

# posted by

Anonymous : August 08, 2005 6:29 PM

Well, because a zero-length array does not allocate much memory (or any, for what it's worth)! I mean, you know what that 1 stands for, right?

# posted by

behdad : August 08, 2005 6:48 PM

I am the first Anonymous (Paolo, an italian who found this article looking for random things ;-) ).
I've tried for curiousity to use the zero sized array for this trick, and it seems to work. Size of the variable it's said to be 0, by the sizeof operator...
By the way, I agree that obscuring the pass by variable it's bad in C, in this case, because you can't do the same trick with any other type. In C++ instead they uses references to do this , and, since it works with any data type, the programmer it's (should be) more careful about doing assumptions.

# posted by

Anonymous : August 09, 2005 8:33 AM

Passing-by-reference isn't just syntactically different than passing-by-pointer. References may only refer to a valid object, and cannot be made to refer to a different object at runtime. Pointers, on the other hand, can be changed to point at any arbitrary memory address (including invalid ones). So while they're often used to achieve the same thing (avoiding a costly pass-by-value copy of a function argument), they have subtle differences. I'm sure you knew this already, but in case anybody didn't...

# posted by

Anonymous : April 01, 2006 10:30 AM

Almost true. In Java you still have null references, right?

# posted by

behdad : April 01, 2006 2:52 PM

The struct is unnecessary. This will do too and is even simpler:

typedef int the_type[1];

void setter(the_type i, int v)
{
i[0] = v;
}

int getter(the_type i)
{
returni[0];
}

# posted by

Anonymous : December 17, 2007 9:29 AM

roel, that has the problem raised here.

# posted by

behdad : December 17, 2007 11:46 AM

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