Using pointers with Struct

[JamesD]

I was trying to make a program that will use a struct to take in 3 employee ID's, first and last names. But have to use a pointer for passing to the function. Here is the code I have, sorry as I am new to programming so there are a few flaws. Any tips would be appreciated.

#include <stdio.h>
struct employee
{
	char firstname[40];
	char lastname[40];
	int id;
};
typedef struct employee Employee[3];

/* Input the employee data interactively from the keyboard */
void InputEmployeeRecord(Employee *ptrEmployee);

void main()
{
    Employee e1;
    InputEmployeeRecord(&e1);
}

void InputEmployeeRecord(Employee *ptrEmployee)
{
	int i=0;
	for(i=0;i<3;i++)
	{
		printf("Enter the employee ID")
		scanf("%d", Employee.id);
		printf("Enter the employee first name")
		gets(firstname);
		printf("Enter the employee last name")
		gets(lastname);
	}
}

[Subsonics]

The best approach is to make things incrementally, and test. Instead of writing everything at once, then compile. You have the data type Employee, which is an array of 3 struct employee. You then pass a pointer to this array of struct employee to your function.

This means that you either need to cast Employee inside InputEmployeeRecord or use the -> convenience syntax. And, Employee is an array, so you need an index as well. Which means if you need to refer to id, you have to do like this: ptrEmployee[i]->id

Note that, Employee is a type you can not use that by itself, but instead *ptrEmployee that you have declared being a pointer of type Employee.

You shouldn't use gets, but fgets, and main should return 0 and have int as a return type.

[Salem]

> typedef struct employee Employee[3];
Typedefs which just make an array out of a type, or make a pointer out of a type don't add a lot of value, but add a hell of a lot of confusion.

Without this, your code looks like this

#include <stdio.h>
struct employee
{
	char firstname[40];
	char lastname[40];
	int id;
};
typedef struct employee Employee;

/* Input the employee data interactively from the keyboard */
void InputEmployeeRecord(Employee *ptrEmployee);

int main()  /*!! yes, really, it is int */
{
    Employee e1[3];
    InputEmployeeRecord(e1);
    return 0;
}

void InputEmployeeRecord(Employee *ptrEmployee)
{
	int i=0;
	for(i=0;i<3;i++)
	{
		/// input code, not using gets()
	}
}

[JamesD]

Why not use gets? Does that not run as well, just curious? And I will start trying to fix up what I have, Ill let you know how it goes. Also why is it better to have main as an int not void?

[rags_to_riches]

Cprogramming.com FAQ > Why gets() is bad / Buffer Overflows
http://www.gidnetwork.com/b-56.html

[JamesD]

Thanks for that, I really hadn't known before so thanks for the advice, I changed up the program but when I run I get multiple errors still, any advice on what I am doing wrong? I am pretty sure its with the pointer in the function and with the word employee inside the function.

#include <stdio.h>
struct employee
{
	char firstname[40];
	char lastname[40];
	int id;
};
typedef struct employee Employee;

void InputEmployeeRecord(Employee *ptrEmployee);

int main()
{
    Employee e1[3];
    InputEmployeeRecord(&e1);
    return 0;
}

void InputEmployeeRecord(Employee *ptrEmployee)
{
	int i=0;
	for(i=0;i<3;i++)
	{
		printf("Enter the employee ID");
		scanf("%d",Employee[i].id);
		printf("Enter the employee first name");
		scanf("%s",Employee[i].firstname);
		printf("Enter the employee last name");
		scanf("%s",Employee[i].lastname);
	}
}

[Clairvoyant1332]

When passing an array to a function, it automatically gets converted to a pointer. So when you call InputEmployeeRecord(&e1) you're actually passing a pointer to a pointer. Get rid of the dereference and you should be fine.

[JamesD]

I actually solved it, and just saw your post now Clairvoyant1332, this is what I ended up with:

#include <stdio.h>
struct employee
{
	char firstname[40];
	char lastname[40];
	int id;
};
typedef struct employee Employee;

void InputEmployeeRecord(Employee *ptrEmployee);

int main()
{
    Employee e1[3];
    InputEmployeeRecord(e1);
    return 0;
}

void InputEmployeeRecord(Employee *ptrEmployee)
{
	int i=0;
	for(i=0;i<3;i++)
	{
		printf("Enter the employee ID:\n");
		scanf("%d",&ptrEmployee[i].id);
		printf("Enter the employee first name:\n");
		scanf("%s",ptrEmployee[i].firstname);
		printf("Enter the employee last name:\n");
		scanf("%s",ptrEmployee[i].lastname);
	}
}

guess I just had to fiddle around, and use your tips. Thanks everyone for helping, really appreciate it.

[rags_to_riches]

scanf("%d",Employee[i].id);

You need to provide the address of a variable when using scanf so that it has a memory location to read into. Use

scanf("%d",&Employee[i].id);

[JamesD]

Yea, I forgot about that before, good call there, and thanks again.

[Salem]

Actually, &e1 (in this case) is a pointer to an array, not a pointer to a pointer.

To pass &e1, the function prototype would need to be.
void InputEmployeeRecord(Employee (*ptrEmployee)[3] );

[Subsonics]

Your actually no better off, using scanf("%s", *p) than gets() if you don't restrict how many characters scanf() can read.

In your case you can do like this if you want to use it: scanf("%39s", *p);
That will leave room for your string and terminating null character.

[JamesD]

I never knew about being able to restrict with scanf, thanks for showing me that, and I think the e1 works, I will have to check else I can change that too, this is a really helpful board . It all works, I built another function to print it out and verify too, I like how you don't give the answer only assist, much better for learning.