C structure; overcoming passing a string to a char - c

Question

I am trying to print a simple structure, which does not work because I am trying to pass a string into a char (I know this is illegal). Yet I cannot figure out how to overcome this. I know there's a way with strcpy and without strcpy, how can I correct this? I have searched the forum and found similar questions, but was unable to apply them to my problem.
#include <stdio.h>
struct article
{
int number;
int quantity;
char description[21];
};
void print(struct article *p)
{
printf("Article Number: %d\nQuantity: %d\nDescription: %s", p->number, p->quantity, p->description);
}
int main()
{
struct article *mydata;
mydata->number = 333;
mydata->quantity = 465;
mydata->description = "Wall Street Journal\n";
print(mydata);
return 0;
}

Answer 1

struct article mydata = {
333,
465,
"Wall Street Journal\n"
};
print(&mydata);
struct article mydata;
mydata.number = 333;
mydata.quantity = 465;
strcpy(mydata.description, "Wall Street Journal\n");
print(&mydata);

Answer 2

First off, you are declaring a pointer to a struct article. You need to allocate memory for that pointer. It won't be allocated on the stack automatically for you. Using malloc you allocate dynamic memory to the heap.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
struct article *mydata = malloc(sizeof(struct article));
if(article == NULL) {
exit(1);
}
mydata->number = 333;
mydata->quantity = 465;
char *description = "Wall Street Journal\n";
size_t len = strlen(description);
memcpy(mydata->description, description, len);
mydata->description[len] = '\0';
print(mydata);
free(mydata);
return 0;
}

Answer 3

You can also do this. But why do you need alternative ways if you have a function that can do the job for you.
char mydescription[]="Wall Street Journal\n";
int count =0 ;
for( count =0 ; mydescription[count] != '\0' ; count++)
{
mydata->description[count] = mydescription[count] ;
}