Initializing and accessing char array in struct

Question

I cannot seem to find a way to initialize a struct without getting segmentation fault . Here is the assignment `

int id = 5;
// scanf ("%10d", &id);
printf("Please give an name\n");
char *tmpName = (char*)malloc(MAXSTRING * sizeof(char));
fgets(tmpName,MAXSTRING,stdin);
student newStudent = (student){ .id = id , .name = tmpName };
printf("%d",newStudent.id);
printf("%s",newStudent.name);

` And here is the struct itself

#define MAXSTRING  256
typedef struct{
    char *name;
    int id;
}student;

I can successfully initialize the struct , but I cannot get access to the name variable , any thoughts?

EDIT : Answers submitted at the time offered nothing , the problem was the way the struct was initialized

char tmpName[MAXSTRING] =  {0};
scanf("%s",tmpName);
student newStudent = { .id = id };
strcpy(newStudent.name,tmpName );

This block fixed the issue , will close the topic.

Answer 1

You define a pointer

char *tmpName;

you do nothing to make it actually point to some useable space, especially there is no malloc() or similar.
The you have scanf() (if successful...) write to the pointer, but a string, not any meaningful address. I.e. if that string is not extremely short, you certainly write beyond.

scanf("%s",&tmpName);

Do not be surprised by segfaults, be surprised if there are none.
Later on you then read from where that weird non-pointer points to...

To solve, insert a malloc() after the pointer definition. Alternatively use fixed array of char as an input buffer.
Use the pointer itself, not its address, in scanf() to write the string into the malloced space. (Or the array identifier.)

The rest is the typical set of problems with having enough space, failing to scan, without checking return value etc. Here is a great set of basic hints for getting input right:
How to read / parse input in C? The FAQ

Answer 2

When you declare char name[MAXSTRING], you are telling your compiler that it will always have the exact length of MAXSTRING.

When the compiler hears that, he will most likely replace your string with a lot of static char members.

#define MAXSTRING  4
typedef struct{
    char name[MAXSTRING];
    int id;
}student;

Would be extended and compiled more like so :

#define MAXSTRING  4
typedef struct{
    char name0;
    char name1;
    char name2;
    char name3;
    int id;
}student;

Although you would still need to access those using the syntax name[index], as you declared an array, the array pointer is in fact nowhere to be found. So you can't directly assign it.

This is basically what happens with any fixed-length array, they won't allow you to asign them since their content is directly stored in the stack or in the datastructure enclosing it.

You would need to declare.

typedef struct{
    char *name;
    int id;
}student;

To get a char array pointer of variable length that you can assignate.

---

That should answer your question, now, as stated by someone in your comment section, there are other things wrong in your code and even with the correct structure, you will probably be unnable to do what you want since scanf is probably returning an arror right now. ^^'

Try allocating the memory beforehand doing so :

char *yourstring = (char*)malloc(MAXSTRING * sizeof(char));

And of course, don't forget to free when you don't need either the string or your student structure :

free(yourstring);

Then again, there's another small issue with your assignement since you do something like this :

.name = *yourstring

While using pointers, prefixing them with * will access the value at the pointed address.

Supposing char *yourstring = "I love cats.";
Doing .name = *yourstring will result in .name being equal to 'I' character, which is first in the array pointed to by yourstring. (Most compiler would complain and ask you to cast.. do not do that.)

So what you really need to do is to assign the pointer to your .name array pointer.

.name = yourstring