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

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;
}

Pointers work like pointers anywhere, it doesn't matter if you have it as a separate local variable, argument, or part as a structure. — Some programmer dude, Jul 05 '20 at 13:44
And for `scanf` with the `%s` format, it expects a pointer to the first element of an array of characters, of type `char *`. When you use the address-of operator you get a pointer *to the pointer*, and its type is `char **`. And same with `printf` and the `%s` format, it expects a pointer to the first element of an array of characters. — Some programmer dude, Jul 05 '20 at 13:45
To be brutally honest, it seems like you have skipped way to much of your text-book, reading a really bad tutorial, or skipped to many classes. Start over, and if you don't have a book, *get one*, you can't learn a language (programming or spoken) by just guessing. — Some programmer dude, Jul 05 '20 at 13:46

Vlad from Moscow · Answer 1 · 2020-07-06T12:50:47.447

0

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.

edited Jul 06 '20 at 12:50

answered Jul 05 '20 at 13:47

Vlad from Moscow

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RobertS supports Monica Cellio · Answer 2 · 2020-07-05T18:12:57.897

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

@StoryTeller-UnslanderMonica Thank you! Edited. – RobertS supports Monica Cellio Jul 05 '20 at 18:15 — RobertS supports Monica Cellio, Jul 05 '20 at 18:15

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

2 Answers2