How is an string being hold by a pointer and not by an character array in C?

Question

This code works perfectly; as you can see here name is a pointer. When I give a value to it, it is stored perfectly, but when I change the pointer to array the code breaks. Why do I need a pointer to store a string?

#include<stdio.h>
#include<stdlib.h>

struct student{
    char *name;
    int roll;
    int clas;
}*ptr,stu;

void main(){
    int i;
    int n;
    stu.roll=2;
    stu.name = "sum";
    printf("%d %s",stu.roll,stu.name);
}

But this doesn't:

#include<stdio.h>
#include<stdlib.h>
struct student{
    char name[10];
    int roll;
    int clas;
}*ptr,stu;

void main(){
    int i;
    int n;

    stu.roll=2;
    stu.name = "sum";
    printf("%d %s",stu.roll,stu.name);
}

Answer 1

You cannot assign to arrays. This doesn't have a meaning in C.

You want:

strcpy(stu.name,"sum");

Answer 2

You don't need to use a pointer to store strings. Arrays are pointers are two valid ways to deal with strings, and they are some way mutually bound with each other. You just need to understand what an array is and how are characters stored in them so that we can call them "strings".

---

First thing we need to remember: what is a pointer (to a type T )?

• It the address where data of type T is stored

Now, what is an array `T var[N]?

• It is a sequence of N elements of the same type T stored contiguously in memory.
• var is the name of the array, and represents the address of its first element of the array. It is evaluated like a pointer though it's not a pointer .

Almost there. What is a string?

• it is an array containing elements of type char terminated by a special nul-terminator character (`'\\0')

---

So a string IS an array, but like every array can be evaluated as a pointer to its first element. And every pointer to char is a string if it is terminated by the null character, and be accessed with the array syntax:

char * str = "hello"; // contains 'h', 'e', 'l', 'l', 'o', '\0'
                      // cannot be modified because it points to a constant area of memory

printf ("%c %c\n", str[1], str[4]); // prints "e o"

/********/

char str2[] = "hello"; // same contents as above, but thi one can be modified

---

Note : the assignment

stu.name = "sum";

is invalid because name is an array field of the struct student struct. As explained above an array is not a pointer, and one of the main differences is that it cannot be assigned (ie cannot be an lvalue in an assignment action). It will raise a compilation error.

The correct action is copying the data into the array using strcpy() function.