simple pointer to structures

Question

I am trying to write simple program to collect data for certain number of students and output it in the end. After I enter data for one student, my program crashes.

Here is the code:

#include <stdio.h>
#include <stdlib.h>
typedef struct Student Student;
struct Student{
    char name[20];
    char lastname[20];
    int age;
};

main() {
    int i;
    int n;
    scanf("%d",&n);


    Student *pStudents = NULL;
    pStudents = (Student*)malloc(n*sizeof(Student));

    for(i=0;i<n;i++) {
        printf("Enter the students name: \n");
        scanf("%s",(pStudents+i)->name);
        printf("Enter lastname: \n");
        scanf("%s",(pStudents+i)->lastname);
        printf("Enter age: \n");
        scanf("%d",(pStudents+i)->age);
    }

    for(i=0;i<n;i++) {
        printf("%s",(pStudents+i)->name);
        printf("%s",(pStudents+i)->lastname);
        printf("%d",(pStudents+i)->age);
    }

}

Thanks in advance.

The error was obviously the scanf but why are you using (pStudents+i) instead of pStudents[i]? It would at least show that pStudents is an array. — Maxime Chéramy, Jun 23 '13 at 15:47
Right. I forgot pointer is same thing as array. I am still new to pointers, but this certainly makes the code more easier to read. — Jay, Jun 23 '13 at 15:53
Assigning `NULL` to `pStudents` it totally redundant; the return value of `malloc()` overwrites is immediately. — meaning-matters, Jun 23 '13 at 15:56
Also, don't cast the return of the `malloc()`. http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc — Maxime Chéramy, Jun 23 '13 at 15:57

score 5 · Accepted Answer · answered Jun 23 '13 at 15:42

5

   scanf("%d",(pStudents+i)->age);

the argument of scanf must be of a pointer type.

Change (pStudents+i)->age to &(pStudents+i)->age.

answered Jun 23 '13 at 15:42

ouah

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Works! SO I assume -> operator is same as . but also dereferences ? – Jay Jun 23 '13 at 15:49
@Jay you use `.` for accessing a member of a struct variable (e.g. `struct Student variable`), and `->` when you're having a pointer to a struct (e.g. `struct Student *pointerVariable`). – meaning-matters Jun 23 '13 at 15:52
1

@Jay `s->a` is syntactic sugar for `(*s).a` – ouah Jun 23 '13 at 15:53
`pStudents` is a pointer to the first Student structure. `pStudents+i` is the address to the i-th element of your table. `(pStudents+i)->age` is the variable you want to modify. However, scanf require the address of that variable, that's why you need that extra `&`. – Maxime Chéramy Jun 23 '13 at 15:53

simple pointer to structures

1 Answers1