How do pointers in the structure work in the following code? - c

Question

I wanted to use pointer inside a struct to store 2 names the code works fine it stores and display the names as I want them to but I am confused about the role of the & operator in the scanf and printf statements when I remove the & operator from printf the names don't display. I don't know how its working.
#include <stdio.h>
int main()
{
struct name
{
char *p;
}n1,n2;
printf("ENTER NAME1\n");
scanf("%s",&n1.p);
printf("\nENTER NAME2\n");
scanf("%s",&n2.p);
printf("ENTERED NAMES ARE %s , %s",&n1.p,&n2.p);
return 0;
}

Answer 1

the code works fine
The code has undefined behavior because in these calls
scanf("%s",&n1.p);
scanf("%s",&n2.p);
expressions &n1.p and &n2.p have the incorrect type char ** and are overwritten.
You need to allocate memory where you are going to read strings.

Answer 2

printf("ENTER NAME1\n");
scanf("%s",&n1.p);
printf("\nENTER NAME2\n");
scanf("%s",&n2.p);
Where do you want to store the strings with the names into?
Into the object of the pointer p inside of the respective structure itself?
This is invalid and illegal. Pointers are made to point anywhere, not to be assigned with a value of an object they do point to themselves.
Thus to answer your question:
"How do pointers in the structure work in the following code?"
They don't work at all. Your code is a mess of undefined behavior.
You need to allocate memory first, where p is pointing to, f.e. by using malloc():
n1.p = malloc(sizeof(char) * 20); // Allocates memory for a string up to 19 characters.
if (!n1.p)
{
fputs("Error at allocation!", stderr);
return EXIT_FAILURE;
}
n2.p = malloc(sizeof(char) * 20); // Allocates memory for a string up to 19 characters.
if (!n2.p)
{
fputs("Error at allocation!", stderr);
return EXIT_FAILURE;
}
and then pass the argument to scanf() properly:
printf("ENTER NAME1\n");
scanf("%s", n1.p);
printf("\nENTER NAME2\n");
scanf("%s", n2.p);
(Note n1.p instead of &n1.p. Same goes for n2.p)
Or better:
printf("ENTER NAME1\n");
if (scanf("%s", n1.p) != 1)
{
fputs("Error at input!", stderr);
return EXIT_FAILURE;
}
printf("\nENTER NAME2\n");
if (scanf("%s", n2.p) != 1)
{
fputs("Error at input!", stderr);
return EXIT_FAILURE;
}
to ensure safety at input errors.
Furthermore the %s conversion specifier expects an argument of type char *, not pointer to char * (char **) as it is with &n1.p and &n2.p.
To provide an argument of wrong type invokes undefined behavior, too.
Just use n1.p and n2.p.
Also this printf() call is wrong:
printf("ENTERED NAMES ARE %s , %s", &n1.p,&n2.p);
You need to omit the & too:
printf("ENTERED NAMES ARE %s , %s", n1.p, n2.p);
All in one:
#include <stdio.h>
#include <stdlib.h>
int main (void)
{
struct name
{
char *p;
} n1, n2;
n1.p = malloc(sizeof(char) * 20); // Allocates memory for a string up to 19 characters.
if (!n1.p)
{
fputs("Error at allocation!", stderr);
return EXIT_FAILURE;
}
n2.p = malloc(sizeof(char) * 20); // Allocates memory for a string up to 19 characters.
if (!n2.p)
{
fputs("Error at allocation!", stderr);
return EXIT_FAILURE;
}
printf("ENTER NAME1\n");
if (scanf("%s", n1.p) != 1)
{
fputs("Error at input!", stderr);
return EXIT_FAILUE;
}
printf("ENTER NAME2\n");
if (scanf("%s", n2.p) != 1)
{
fputs("Error at input!", stderr);
return EXIT_FAILURE;
}
printf("ENTERED NAMES ARE %s , %s", n1.p, n2.p);
free(n1.p):
free(n2.p);
return 0;
}
Execution:
ENTER NAME1
Peter
ENTER NAME2
Maria
ENTERED NAMES ARE Peter , Maria