struct node warning

Code:

---

#include <stdio.h>

typedef struct{
        struct name *studentName;
}student;


typedef struct{
        char *firstName;
        char *secondName;
}name;


int main(void){
        student *a;
        name *b;
       
        b=(name*)malloc(sizeof(name));
        a=(student*)malloc(sizeof(student));
       
        b->firstName="david";
        b->secondName="becks";

        a->studentName=b;  //warning line
}

---

Hi,i wrote above a simple struct . the student struct has a link to another name struct to store the 1st and 2nd name.

The code works but i get a warning at the above stated line.
warning: incompatible types -from 'name*' to 'name *'

I hope someone can correct my error.

much apreciated.
Thanks

OK, here's your main problem:

Code:

---

struct name *studentName;

---

Where is "struct name" defined or declared? The answer might surprise you: It hasn't.

Later on you typedef an anonymous struct to be known as name.

Solution? Re-arrange your structs:

Code:

---

typedef struct{
        char *firstName;
        char *secondName;
}name;

typedef struct{
        name *studentName;
}student;

---

Otherwise:

• Actually return a value from main(). According to popular convention, 0 is used to signify success, while non-zero means failure of some sort.
• Check if malloc() fails before you use dynamically allocated memory.
• Include stdlib.h if you use malloc(). If you had warnings turned up, you probably would have been alerted that you missed it:
  
  
  Code:

  ---

  : warning: implicit declaration of function `malloc'

  ---

struct name *studentName is not the same thing as name *studentName. I don't know why it didn't emit an error for an undeclared struct though.

Code:

---

#include <stdio.h>
#include <stdlib.h> /* for malloc */

typedef struct{
        char *firstName;
        char *secondName;
}name;


typedef struct{
        name *studentName;
}student;



int main(void){
        student *a;
        name *b;
       
        b=(name*)malloc(sizeof(name));
        a=(student*)malloc(sizeof(student));
       
        b->firstName="david";
        b->secondName="becks";

        a->studentName=b;
        return 0;
}

---

thank you for your help.I have then proceeded to make it into a list of students.I get an error again.I dont know what wrongs

Code:

---

#include <stdio.h>
#include <stdlib.h> /* for malloc */

typedef struct student* studentPtr;

typedef struct{
        char *firstName;
        char *secondName;
}name;


typedef struct{
        name *studentName;
        studentPtr nextStudent;
}student;



int main(void){
        name *a,*b;
        student *c,*d;
       
        a=(name*)malloc(sizeof(name));
        b=(name*)malloc(sizeof(name));
        c=(student*)malloc(sizeof(student));
        d=(student*)malloc(sizeof(student));
       
        a->firstName="david";        a->secondName="becks";
        b->firstName="Eric";        b->secondName="Olmado";

        c->studentName=a;
        d->studentName=b;

        c->nextStudent=d;  /*warning line*/
        return 0;
}

---

i get the warning of incompatible type.
tried hard to figure it out though.maybe my basic is wrong

You're doing the same wrong thing.

studentPtr is struct student *.
d is of type student *.
struct student * is not not not the same thing as student *.

Until you get this through your head, you're going to continue making the same mistakes. Instead of just copying the code I provided in my first post, go through my explanation and try to understand it. That should make it more clearer as to why your above code is wrong.

you did this:

Code:

---

typedef struct student* studentPtr;

---

before this:

Code:

---

typedef struct{
        name *studentName;
        studentPtr nextStudent;
}student;

---

the typedef doesn't know what struct student is, so it has problems.
you have to define the types that you use before you use them. this includes other syntaxes like when you use typedef.


normally, in C, you have to prepend a 'struct' keyword in front of struct things because that's the syntax. You can get around that with using the typedef, but there are some gotchas to that.

Code:

---

struct s
{
    int a;
};

struct s my_s; /*legal*/

struct b{char my_char;};

b my_b;/*not legal*/

typedef struct b b;

b my_b;/*now legal*/

typedef struct {char my_char;}c;

struct c my_c;/*not legal*/
c my_c;/*legal*/

typedef struct d{char my_char;}d;

struct d my_d;/*legal*/
d my_d;/*legal*/

---

and just to throw you off:

Code:

---

struct e {
    int my_int;
} my_instance;

my_instance i; /*not legal*/

int main()
{
    struct e i; /*legal*/

    i.my_int = 234; /*legal */
    my_instance.my_int = 3254;/*legal*/

    return 0;
}

---

maybe just put

Code:

---

typedef student* studentPtr;

---

after

Code:

---

typedef struct{
        name *studentName;
        studentPtr nextStudent;
}student;

---