Passing structure by reference or pointer?

Given the structure below:-

Code:

---

typedef struct {
 int a;
 word b;
} process;

---

By reference:-

Code:

---

main()
{
 process work;
 /* init structure variable */
 ...
 do_something( &work );
 ...
}

---

By pointer:-

Code:

---

main()
{
 process work;
 process *ptr_work = &work;

 /* init structure variable */
 ...
 do_something( ptr_work );
 ...
}

---

Which would be an optimal performance approach if I want to pass a structure into a function?

Also, some reading up here and there came across to a site that says "by reference" is a C++ solution (see http://cplus.about.com/od/cprogrammi.../aa010402a.htm ). I have this impression that it is available in C as well, so the other question I have is if it is really a C++ approach, does that mean most, if not all, C compilers would give me error if I use by reference method?

Both ways you've shown are passing by pointer...

Here's how to declare a function by reference:

Code:

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void foo(int& a); // notice the amperstand

---

Its parameters are passed just like normal variables.

PS: The first solution seems a lot better to me.

Passing by reference is only allowed in C++. Not under C.
Passing a pointer is not the same as passing a reference. It's important to understand the difference because it is a source of nasty errors.

When you pass a pointer, you are still passing by value. However, it's not the object that is being copied, but the pointer that will hold a local copy inside the function.

Since C doesn't allow passing by reference, all parameters in C hold copies of the arguments passed.

As for your struct question. Passing by pointer will be a better option on most circumstances. The struct object will not be copied. Just the pointer. For large structs this can be beneficial.

Both solutions will have the same effect, the first one will probably save you a couple of bytes worth of memory; it is also easier to follow.

No. Both solutions are not the same. Try changing a reference address, for instance.

Besides this is a C forum. Passing by reference is not recognized by the compiler.

>Try changing a reference address, for instance.

Well, the OP didn't try to; so as it stands they are the same. Either you pass the address of a variable, or you pass a pointer which holds the address of the variable - both will do the same thing.

Of course, if you want to start referencing multiple structures of the same type, a pointer is obviously going to be a better solution.

Ah okie..i confused a pointer method as pass by reference. Thanks for higlighting that out.

And thanks for the inputs.

There is a way to simulate a pass by reference in C, and that is to pass a pointer to a pointer. A demonstration:

Code:

---

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

struct cat {
        char name[32];
        char bday[12];
};

void foo(struct cat ** me) {
        printf("%s is located at %p in foo\n", (*me)->name, *me);
}

int main(void) {
        struct cat *gargamel = malloc(sizeof(struct cat));
        strncpy(gargamel->name, "Gargamel", 32);
        strncpy(gargamel->bday, "4/7/2000", 12);
        printf("%s is located at %p in main\n", gargamel->name, gargamel);
        foo(&gargamel);
        return 0;
}

Owner@PAVILION ~
$ ./a
Gargamel is located at 003D3DA8 in main
Gargamel is located at 003D3DA8 in foo

Owner@PAVILION ~
$

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As you can see, no temporary is made. I like cats.

thank you whitecats, err... i mean whiteflags.... :)

the example that you showed is exactly what i was looking for...