Dynamically allocation memory for array of structs in c

Question

I have two structures as shown below

struct server{
    // some members
};

struct msg{
    struct server* servers;
};

Then I do this.

struct msg msg1;
struct server s1,s2;

msg1.servers = (struct server *)malloc(2*sizeof(struct server));
msg1.servers[0] = &s1; // compilation error 
msg1.servers[1] = &s2;  // compilation error

This code does not compile and giving the following error : incompatible types when assigning to type ‘struct server’ from type ‘struct server *’.

What am I doing wrong?

JaredPar · Accepted Answer · 2012-04-19 18:33:03Z

3

The problem here is that the expression msg1.servers[0] producets a struct server but you're providing a struct server* (a pointer type vs a non-pointer type). There are two ways to fix this

The first is to simply provide the struct server instances by value as the code expects

msg1.servers[0] = s1;
msg1.servers[1] = s2;

This will work if struct server is a type that behaves properly when copied around.

The second is necessary if you want to continue using struct server* in the struct msg. In this case you need a double pointer to store the server pointers. And you need to adjust your malloc statement appropriately

struct msg{
    struct server** servers;
};

msg1.servers = malloc(2*sizeof(struct server*));
msg1.servers[0] = &s1; 
msg1.servers[1] = &s2;

answered Apr 19, 2012 at 18:33

JaredPar

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Comments

Dave · Accepted Answer · 2012-04-19 18:32:36Z

2

You don't need the ampersands:

msg1.servers[0] = s1;
msg1.servers[1] = s2;

Saying &s1 would give you a pointer to a struct server, but msg1.servers[0] is a single element in the array you just allocated.

answered Apr 19, 2012 at 18:32

Dave

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Comments

Praetorian · Accepted Answer · 2012-04-19 18:36:01Z

1

msg.servers is a pointer to the type server. When you use msg.servers[0] you're dereferencing the pointer, so its type is now server, not server *, obviously you cannot assign the address of a server instance to it.

You probably want the following:

struct msg{
  struct server** servers;
};

struct msg msg1;
struct server s1,s2;

msg1.servers = malloc(2 * sizeof(struct server *));
msg1.servers[0] = &s1;
msg1.servers[1] = &s2;

answered Apr 19, 2012 at 18:36

Praetorian

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Comments

John Bode · Accepted Answer · 2012-04-19 18:34:52Z

0

msg1.servers[0] = &s1;

An array subscript implicitly dereferences the pointer (a[i] == *(a + i), so the type of the expression msg1.servers[0] is struct server, not struct server *.

Change your struct msg definition to

struct msg {
  struct server **servers; // servers will be an array of pointers
};

and the malloc call to

msg1.servers = malloc(2 * sizeof *msg1.servers);

and then the assignments will work.

EDIT

Or, like everyone else is pointing out, drop the & from the assignment. It depends on what you intend servers to represent; is it an array of struct server, or an array of pointers to struct server?

answered Apr 19, 2012 at 18:34

John Bode

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Comments

Sangeeth Saravanaraj · Accepted Answer · 2012-04-19 18:37:07Z

0

You need a array of pointers

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

int main(void)
{
    struct server{
        // some members
    };  

    struct msg{
        struct server ** servers;
    };  

    struct msg msg1;
    struct server s1,s2;

    if ((msg1.servers = malloc(2 * sizeof(struct server *))) == NULL) {
        printf("unable to allocate memory \n");
        return -1; 
    }   
    msg1.servers[0] = &s1; // compilation error 
    msg1.servers[1] = &s2;  // compilation error

    free(msg1.servers);

    return 0;
}

answered Apr 19, 2012 at 18:37

Sangeeth Saravanaraj

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