How do you assign a variable in one of the member construct in C?

Question

I used the brackets after the name variable, but I still do not under stand why I am getting error messages. I know that C does not support string, but an array could be used to store string values:
Here is my code:

#include <stdio.h>
struct Student{
char name[12];
int age;

};

int main(){
struct Student s1;
struct Student s2;

s1.name[12] = "Shubham";
s1.age = 21;

s2.name = "Gargi";
s2.age = 13;

printf("\nName: %s \n Age: %d ",s1.name,s1.age);
printf("\nName: %s \n Age: %d ",s2.name,s2.age);
}

Terminal:

ERROR!
gcc /tmp/65wVAJrLIS.c -lm
/tmp/65wVAJrLIS.c: In function 'main':
/tmp/65wVAJrLIS.c:13:17: warning: assignment to 'char' from 'char *' makes integer from pointer without a cast [-Wint-conversion]
13 | s1.name[12] = "Shubham";
| ^
/tmp/65wVAJrLIS.c:16:13: error: assignment to expression with array type
16 | s2.name = "Gargi";
|

Answer 1

Shubham Ambekar just use a char* for name.

#include <stdio.h>
struct Student{
char* name;
int age;
};

int main(){
struct Student s1;
struct Student s2;

s1.name = "Shubham";
s1.age = 21;

s2.name = "Gargi";
s2.age = 13;

printf("\nName: %s \n Age: %d ",s1.name,s1.age);
printf("\nName: %s \n Age: %d ",s2.name,s2.age);
}

Answer 2

A char array can (and indeed must) be used to store strings, but you can't naively assign strings to an array with =, any more than you could do that with an array of numbers.

You either need to assign it char-by-char yourself, or like Codenath shows, #include string.h and let strcpy() do that work for you behind the scenes.