[SOLVED] C: How do you declare a function in advance?

stf92 · 12-14-2014, 10:18 AM

I know how to do it in the following case:

Code:

int f1(char *);

int f1(char *a1){
/* BODY OF THE FUNCTION */
}

But what about this case?

Code:

int f2(char *a1[100]);{
/* BODY OF THE FUNCTION */
}

How would you declare it?

ndc85430 · 12-14-2014, 11:11 AM

I don't understand your question. You've provided a prototype for your function f2(), so what else do you need, besides the implementation, i.e.

Code:

int f2(char *a1[100]) {
 // Code here
}

stf92 · 12-14-2014, 11:24 AM

Sorry. That was meant to be the function itself. I'll will edit the post.

ndc85430 · 12-14-2014, 11:44 AM

But your prototype

Code:

int f2(char *a1[100]);

looks fine to me. Why do you think there's a problem with it?

stf92 · 12-14-2014, 11:52 AM

But if you look at the first prototype in post #1, you'll see there is no variable name there. I infer I could save the variable name in the second too, hence.

rknichols · 12-14-2014, 12:21 PM

The variable name in a function prototype is meaningless, but you can put one there if you like. The name (or in your case, even the array size) does not need to match what appears in the actual function definition. Only the type is important.

stf92 · 12-14-2014, 12:37 PM

But

Code:

bill@juan:~/TIMER/arrays$ cat u01.c
#include <stdlib.h>
#include <stdio.h>
#include <string.h>

int juan(char *);
int juan(char* a[10]){
        /* The first dimension is actually ignored by the compiler. So
         *          * int juan(int a[][10]) is equally good. El segundo indice
es el
         *                   * que varia mas rapidamente.
         *                           */
        printf("The values are %s and %s\n", a[0], a[1]);
        return 0;       
}

int main(){
        char* a[10];
        
       return 0;                       
}
bill@juan:~/TIMER/arrays$ gcc u01.c
u01.c:6:5: error: conflicting types for 'juan'
u01.c:5:5: note: previous declaration of 'juan' was here
bill@juan:~/TIMER/arrays$

EDIT: It seems you were right. I put

Code:

int juan(char * []);

and the compiler accepted it.

rknichols · 12-14-2014, 12:45 PM

There is a big difference between a character pointer (char *) and an array of character pointers (char *[]) or (char *a[]).

stf92 · 12-14-2014, 12:57 PM

Thanks.

rtmistler · 12-15-2014, 01:35 PM

If the function in question is above the point where it is used, you do not need to declare a prototype for it. Are you intentions to learn how to say place that function in another file so you can segment a lot of code and be able to call the function from outside file scope? Are your questions related more to the scope of the function as it applies to file versus project?

sundialsvcs · 12-15-2014, 04:15 PM

Quote:

Originally Posted by stf92

But what about this case?

Code:

int f2(char *a1[100]); {
/* BODY OF THE FUNCTION */
}

How would you declare it?

This case is syntactically invalid, AFAIK. Either the declaration of function f2 is followed by a semicolon, which indicates a forward-declaration of that function, or it is immediately followed by the body of the function, in braces, with no semicolon or any other punctuation mark between them. When the "real" definition of the function appears, it must exactly match the previous forward-declaration ... although, fair warning(!), it's pretty much an honor-system. Unlike C++, the C linker does not validate parameter-passing.

Such forward-declarations customarily appear in .h files.

psionl0 · 12-16-2014, 03:41 AM

Quote:

Originally Posted by stf92

EDIT: It seems you were right. I put

Code:

int juan(char * []);

and the compiler accepted it.

You could also have declared int juan(char **);

stf92 · 12-16-2014, 12:16 PM

I think I understand. For every array b, b is a pointer pointing to the first of the elements of the array. So, in

char * a1[100];

a1 points to a1[0]. But according to the declaration, the elements of a1 are pointers themselves. So a1 points to a1[0] which in turn points to the first element of a1[0] (which is char). Hence a1 is pointer to pointer and I may write

char ** a1;

Actually it would be (char *) * a1, so I think the compiler associates from left to right in the expression above.

NevemTeve · 12-17-2014, 02:04 AM

Yes, as function-parameter declaration they are the same thing.

Mind you, as variable declaration they are different: the first allocates 400 bytes (on 32-bit computer) for 100 pointers of type 'char *', the second allocates 4 bytes for 1 pointer of type 'char **'

stf92 · 12-17-2014, 07:36 AM

Thank you so much, NevemTeve.