Passing struct to function

Question

I'm a new C programmer and I wanted to know how I can pass a struct through to a function. I'm getting an error and can't figure out the correct syntax to do it. Here is the code for it....

Struct:

struct student{
    char firstname[30];
    char surname[30];
};

struct student person;

Call:

addStudent(person);

Prototype:

void addStudent(struct student);

and the actual function:

void addStudent(person)
{
    return;
}

Compiler errors:

line 21: warning: dubious tag declaration: struct student
line 223: argument #1 is incompatible with prototype:

Answer 1

This is how to pass the struct by reference. This means that your function can access the struct outside of the function and modify its values. You do this by passing a pointer to the structure to the function.

#include <stdio.h>
/* card structure definition */
struct card
{
    int face; // define pointer face
}; // end structure card

typedef struct card Card ;

/* prototype */
void passByReference(Card *c) ;

int main(void)
{
    Card c ;
    c.face = 1 ;
    Card *cptr = &c ; // pointer to Card c

    printf("The value of c before function passing = %d\n", c.face);
    printf("The value of cptr before function = %d\n",cptr->face);

    passByReference(cptr);

    printf("The value of c after function passing = %d\n", c.face);

    return 0 ; // successfully ran program
}

void passByReference(Card *c)
{
    c->face = 4;
}

This is how you pass the struct by value so that your function receives a copy of the struct and cannot access the exterior structure to modify it. By exterior I mean outside the function.

#include <stdio.h>


/* global card structure definition */
struct card
{
    int face ; // define pointer face
};// end structure card

typedef struct card Card ;

/* function prototypes */
void passByValue(Card c);

int main(void)
{
    Card c ;
    c.face = 1;

    printf("c.face before passByValue() = %d\n", c.face);

    passByValue(c);

    printf("c.face after passByValue() = %d\n",c.face);
    printf("As you can see the value of c did not change\n");
    printf("\nand the Card c inside the function has been destroyed"
        "\n(no longer in memory)");
}


void passByValue(Card c)
{
    c.face = 5;
}

Answer 2

The line function implementation should be:

void addStudent(struct student person) {

}

person is not a type but a variable, you cannot use it as the type of a function parameter.

Also, make sure your struct is defined before the prototype of the function addStudent as the prototype uses it.

Answer 3

When passing a struct to another function, it would usually be better to do as Donnell suggested above and pass it by reference instead.

A very good reason for this is that it makes things easier if you want to make changes that will be reflected when you return to the function that created the instance of it.

Here is an example of the simplest way to do this:

#include <stdio.h>

typedef struct student {
    int age;
} student;

void addStudent(student *s) {
    /* Here we can use the arrow operator (->) to dereference 
       the pointer and access any of it's members: */
    s->age = 10;
}

int main(void) {

    student aStudent = {0};     /* create an instance of the student struct */
    addStudent(&aStudent);      /* pass a pointer to the instance */

    printf("%d", aStudent.age);

    return 0;
}

In this example, the argument for the addStudent() function is a pointer to an instance of a student struct - student *s. In main(), we create an instance of the student struct and then pass a reference to it to our addStudent() function using the reference operator (&).

In the addStudent() function we can make use of the arrow operator (->) to dereference the pointer, and access any of it's members (functionally equivalent to: (*s).age).

Any changes that we make in the addStudent() function will be reflected when we return to main(), because the pointer gave us a reference to where in the memory the instance of the student struct is being stored. This is illustrated by the printf(), which will output "10" in this example.

Had you not passed a reference, you would actually be working with a copy of the struct you passed in to the function, meaning that any changes would not be reflected when you return to main - unless you implemented a way of passing the new version of the struct back to main or something along those lines!

Although pointers may seem off-putting at first, once you get your head around how they work and why they are so handy they become second nature, and you wonder how you ever coped without them!

Answer 4

You need to specify a type on person:

void addStudent(struct student person) {
...
}

Also, you can typedef your struct to avoid having to type struct every time you use it:

typedef struct student{
...
} student_t;

void addStudent(student_t person) {
...
}

Answer 5

Instead of:

void addStudent(person)
{
    return;
}

try this:

void addStudent(student person)
{
    return;
}

Since you have already declared a structure called 'student' you don't necessarily have to specify so in the function implementation as in:

void addStudent(struct student person)
{
    return;
}