C struct

Question

typedef struct Weapon {
int dmg;
char name[];
} Weapon;

////////////

Weapon knife;
knife.dmg = 40; // all good
knife.name = "knife" // expression must be a modifiable value

Help :/

Answer 1

knife.name = malloc(sizeof(char) * 6);
knife.name[5] = '\0';
strcpy(knife.name, "knife");

Or easier

typedef struct Weapon {
int dmg;
char name[50];
};

Weapon knife;
strcpy(knife.name, "knife");

Answer 2

Thanks Just4f
TheWhiteCat💡 even with it didnt work

Answer 3

struct struturename
{
datamember1,
datamember2,
—
—

}variable;

structure is a collection of heterogeneous datatypes

The variables in structures are datamembers

Every structure must have a variable called struture variable

if it is normal variable use dot(.) operator to access structure variables

if it is pointer use arrow(->) opreator to access structure variables

structure with in structure is called nested structure

ex:

struct student
{
int sno;
char sname[30];
};
void main()
{
struct student s;
struct student *p;
clrscr();
printf(“Enter no :”);
scanf(“%d”,&s.sno);
printf(“Enter name :”);
scanf(“%s”,s.sname);
printf(“Number = %d\n”,s.sno);
printf(“Name = %s\n”,s.sname);
printf(“\n”);
printf(“Enter no :”);
scanf(“%d”,&p->sno);
printf(“Enter name :”);
scanf(“%s”,p->sname);
printf(“Number = %d\n”,p->sno);
printf(“Name = %s\n”,p->sname);
getch();
}

Answer 4

You could try the following option:
typedef struct {
int dmg;
char *name;
} Weapon;

/////////////

Weapon knife;
knife.dmg = 40;
knife.name = "knife";

Answer 5

I think you should set the size of name char array => for example char name [20].