return an array of structs

So I have been obsessing over creating a merge sort for linked lists. I can do a merge sort with an int array fine, but there seems to be much less flexability dealing with an array of structs which is giving me a lot of problems. How do you return an array of structs? (After trying it and googling I am not sure if it is possible, strangely). Trying to pass such a list as an argument seems even less possible.

Ideally, I would to do this with just an array of struct pointers, or an array of any kind of pointers but my compiler will not let me cast them into struct pointers in a useful way.

You can not return an array from a function, no matter what.

You can pass in an array and fill it in, or you can return a pointer, or pass in a pointer to pointer and allocate space (and either pass in a pointer to an itneger that gives you the number of items, or return the number of items if you pass a pointer to pointer).

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Mats

Quote:

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Originally Posted by matsp

You can not return an array from a function, no matter what.

You can pass in an array and fill it in, or you can return a pointer, or pass in a pointer to pointer and allocate space (and either pass in a pointer to an itneger that gives you the number of items, or return the number of items if you pass a pointer to pointer).

--
Mats

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If I pass in an array, I have to declare it's length?

Code:

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void testfunc (struct *ray[10]) {

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If I return a pointer (eg, struct *ptr (struct *ray) {) should I allocate for the array it points to inside the function, or before I call it?

Here's an example of the wierdness I'm talking about:

Code:

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

struct Test {
        char *name;
        int age;
} ; 

void testfunc (struct Test *examp) {
        if ((examp=realloc(examp,3*sizeof(*examp)))==NULL) puts ("malloc fail");
        puts("HERE");
        fflush(stdout);
        examp[0].name=malloc(5);
        strcpy(examp[0].name,"five");
        examp[0].age=5;
        examp[1].name=malloc(5);
        strcpy(examp[1].name,"four");
        examp[1].age=4;
        examp[2].name=malloc(5);
        strcpy(examp[2].name,"nine");
        examp[2].age=9;
}

int main () {
        int i=0;
        struct Test *examp=malloc(sizeof(*examp));         
       
        testfunc (examp);
        printf("sizeof *examp:%d sizeof examp:%d\n",sizeof(*examp),sizeof(examp));
        printf("%s is %d\n", examp[0].name,examp[0].age);
        for (i=0;i<3;i++) {puts("X");printf("%s is %d\n", examp[i].name,examp[i].age);}
       
}

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Everything is fine. I am surprised that *examp is 8 bytes while examp is 4 bytes, but whatever (???). However, I am a little worried the realloc doesn't really realloc inside the function, because this causes a segfault:

Code:

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

struct Test {
        char *name;
        int age;
} ; 

void testfunc (struct Test *examp) {
        if ((examp=malloc(3*sizeof(*examp)))==NULL) puts ("malloc fail");
        puts("HERE");
        fflush(stdout);
        examp[0].name=malloc(5);
        strcpy(examp[0].name,"five");
        examp[0].age=5;
        examp[1].name=malloc(5);
        strcpy(examp[1].name,"four");
        examp[1].age=4;
        examp[2].name=malloc(5);
        strcpy(examp[2].name,"nine");
        examp[2].age=9;
}

int main () {
        int i=0;
        struct Test *examp;         
       
        testfunc (examp);
        printf("sizeof *examp:%d sizeof examp:%d\n",sizeof(*examp),sizeof(examp));
        printf("%s is %d\n", examp[0].name,examp[0].age);
        for (i=0;i<3;i++) {puts("X");printf("%s is %d\n", examp[i].name,examp[i].age);}
       
}

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Why can't I malloc the struct array inside another function and return it?

Quote:

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Originally Posted by MK27

Everything is fine. I am surprised that *examp is 8 bytes while examp is 4 bytes, but whatever (???).

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examp is a pointer, which (on a 32-bit system) is 4 bytes in size. *examp is what its points to, which in this case is the size of your structure.

Quote:

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Originally Posted by MK27

Why can't I malloc the struct array inside another function and return it?

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You can. The drawback to this technique is that the callers of your function will have to remember to free() the array.

--
Computer Programming: An Introduction for the Scientifically Inclined

However, if you want to pass in a pointer, to be filled in inside the function, you should pass a pointer to a pointer - just like if you want to change an integer inside a function, you need to pass a pointer to an integer.

--
Mats

What you need or have is a 2d array and you need to properly set aside memory for that as well.

Code:

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int **createMyExamp(int r, int c)
{
  int i;
  int **examp = malloc(r * sizeof(int *));
  for (i = 0 ; i < r; i++)
      examp[i] = malloc (c * sizeof(int));
  return examp;
}

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Code:

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int (*p)[dimension2] = malloc(dimension1 * sizeof(type) + dimension2);

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That you need a p2p to create a 2D array is a myth. All you need is the correct type.

Changed the function to return the structure and this works

Code:

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

typedef struct Test {
        char *name;
        int age;
} ARF ;

int main () {
        int i;
        ARF *examp;

        ARF *testfunc() ;

        examp = testfunc();

        printf("sizeof *examp:%d sizeof examp:%d\n",sizeof(*examp),sizeof(examp));
        printf("%s is %d \n", examp[0].name, examp[0].age );

        for (i=0;i<3;i++) printf("%s is %d\n", examp[i].name,examp[i].age);

        return 0 ;

}

ARF *testfunc()
{
        ARF *bexamp ;
        if (  ( bexamp =  (struct Test *) malloc( 3 * sizeof(ARF) ) ) == NULL)  puts ("malloc fail");

        puts("HERE");
        fflush(stdout);

        bexamp[0].name=(char *) malloc ( (strlen("five")+1) * sizeof(char ) );
        strcpy(bexamp[0].name,"five");
        bexamp[0].age=5;

        bexamp[1].name=(char * ) malloc(5 * sizeof(char ) );
        strcpy(bexamp[1].name,"four");
        bexamp[1].age=4;

        bexamp[2].name=(char * ) malloc(5 * sizeof(char ) );
        strcpy(bexamp[2].name,"nine");
        bexamp[2].age=9;

        return bexamp ;
}

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you should not use null-pointer (after failed malloc)

Quote:

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Originally Posted by Elysia

Code:

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int (*p)[dimension2] = malloc(dimension1 * sizeof(type) + dimension2);

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That you need a p2p to create a 2D array is a myth. All you need is the correct type.

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Sorry to dig this up, but I've been looking into this thread and got some nice info. I have a few doubts regarding the quoted statement, can someone elaborate on it? What is p2p? (In this context, I think its not peer to peer)


I believe this is an answer to , if yes how does it apply?

Quote:

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Code:

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int **createMyExamp(int r, int c)
{
  int i;
  int **examp = malloc(r * sizeof(int *));
  for (i = 0 ; i < r; i++)
      examp[i] = malloc (c * sizeof(int));
  return examp;
}

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Thanks!

Quote:

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Originally Posted by sugarfree

What is p2p?

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A "pointer to a pointer", eg:

Code:

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  int **examp = malloc(r * sizeof(int *));

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haha, silly me. That was easy.

Another question, what is the point of having examp[i] = malloc (c * sizeof(int)); in slingerland3g post? Is this examp[r][c]?

Quote:

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Originally Posted by sugarfree

Is this examp[r][c]?

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Yes.

example[r] is a pointer and needs c*sizeof(int) bytes allocated to it to store an array (row) of ints.