Creating an array of struct (c)

Question

i've created a struct "Employee"

#define MAX_SIZE 20

typedef struct Employee{
    char name[MAX_SIZE];
    int s;
    int e;
} employee_s;

and i need to create an array of 2 employees and ask the user to initialize them, nothing i try seem to work,

void main()
{
    int i, s, e;
    char name[10];
    Employee* e[3];

    *e = (Employee*)malloc(sizeof(Employee)*ARR_SIZE);

    for(i=0; i < 3; i++)
    {
        fflush(stdin);
        puts("Please enter Employee's name:");
        scanf("%c",&name);
        *e[i]->name = name;

        puts("Please enter Employee's salary:");
        scanf("%d",&s);
       *e[i]->s= s;

        puts("Please enter Employee's experience:");
        scanf("%d",&e);
       *e[i]->e=e;

    }
}

p.s: i dont have to use dynamic allocation,
what do i do wrong?

thank you!

score 4 · Accepted Answer · answered Sep 02 '15 at 15:52

4

There are several errors here:

Employee is not a valid type. struct Employee is, and so is employee_s.
e is defined in multiple places
When reading in name, use %s (for a string), not %c (for a char)
Your employee array is defined as an array of pointers. That's probably not what you want. You just need an array. No call to malloc either.
Never fflush(stdin). It's undefined behavior.
In your scanf calls, put a space as the first character in the string. That will allow any newlines to be passed over.

The result:

int main()
{
    int i;
    employee_s e[3];

    for(i=0; i < 3; i++)
    {
        puts("Please enter Employee's name:");
        scanf(" %s",&e[i].name);

        puts("Please enter Employee's salary:");
        scanf(" %d",&e[i].s);

        puts("Please enter Employee's experience:");
        scanf(" %d",&e[i].e);
    }

    for(i=0; i < 3; i++) {
        printf("emp %d: name=%s, sal=%d, exp=%d\n", i, e[i].name, e[i].s, e[i].e);
    }
}

answered Sep 02 '15 at 15:52

dbush

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`&e[i].name` should be `e[i].name`. And those spaces before the format specifier are unnecessary. – Spikatrix Sep 02 '15 at 16:23
@CoolGuy Although generally we don't use `&` while reading in character array but I think there is not any problem in using `&e[i].name` instead of `e[i].name`. Take a look at this [question](http://stackoverflow.com/questions/5406935/reading-a-string-with-scanf) – Hardik Modha Sep 02 '15 at 16:27
thank you very much! one thing i did not understand is why not using fflush(stdin), if i want to recive string it will not take the garbage values in the buffer otherwise? – Coder123 Sep 02 '15 at 16:30
@HardikModha Hmm. But as that answer you linked says, it is wrong and the standard doesn't say anything about it. Also, you get a warning. So it is better to avoid it. – Spikatrix Sep 02 '15 at 16:35
@user3142625 What? Could you clarify? Your comment seems to be unclear for me. – Spikatrix Sep 02 '15 at 16:37
@CoolGuy Yes..We should avoid it. I was just making clear that we can use `&` also. (see my answer I haven't used it! ;) ) – Hardik Modha Sep 02 '15 at 16:39
hmm dbush wrote "Never fflush(stdin). It's undefined behavior." i do not understand why :( – Coder123 Sep 02 '15 at 16:55
@user3142625 The leading space in each `scanf` format string matches zero or more whitespace characters, so any newlines or spaces are skipped over. Also, from the `fflush` manpage: "The function fflush() forces a write of all user-space buffered data for the given *output or update stream* via the stream’s underlying write function." `stdin` is not an output or update stream. – dbush Sep 02 '15 at 17:00

score 2 · Answer 2 · answered Sep 02 '15 at 15:52

You've got your declaration backward. This:

typedef struct Employee{
    char name[MAX_SIZE];
    int s;
    int e;
} employee_s;

declares a type named employee_s to be equivalent to struct Employee, and furthermore declares struct Employee. You appear to want this, instead:

typedef struct employee_s {
    char name[MAX_SIZE];
    int s;
    int e;
} Employee;

In this case you can omit employee_s from that if you wish; perhaps that would be less confusing.

Moreover, you are going about your allocation in a very strange way, especially since you don't require dynamic allocation. Why not just do this:

Employee e[3];

? Then you can (and should) skip malloc() altogether. You will then refer to the members of the array elements via the form e[0].name, etc..

Hardik Modha · Answer 3 · 2015-09-02T16:48:48.473

You can do this easily without dynamic memory allocation as follows.

#include <stdio.h>
#define MAX_SIZE 20
typedef struct Employee{
    char name[MAX_SIZE];
    int s;
    int e;
} employee_alias;
int main(void) {
    int i;
    employee_alias e[3];
    for(i=0; i < 3; i++)
    {
        puts("Please enter Employee's name:");
        scanf("%s",e[i].name);
        puts("Please enter Employee's salary:");
        scanf("%d",&e[i].s);
        puts("Please enter Employee's experience:");
        scanf("%d",&e[i].e);

        printf("Entered Data\n");
        printf("Name : %s\n",e[i].name);
        printf("Salary : %d\n",e[i].s);
        printf("Experience : %d\n",e[i].e);
    }
    return 0;
}

`char name[10];` is unused. So, Remove it – Spikatrix Sep 02 '15 at 16:44 — Spikatrix, Sep 02 '15 at 16:44

Creating an array of struct (c)

3 Answers3