How give value to structures in c?

Question

I'm working with structures in c, but I can not give value to the structures' attributes.

#include <stdio.h>
#include <string.h>

struct Book{
        char name[10];
        int id;
};

int main(){
        char tmp_name[10];
        int tmp_id;
        for(;;){
                struct Book a;
                scanf("%s",tmp_name);
                scanf("%d", tmp_id);
                strcpy(a.name,tmp_name);
                a.id = tmp_id;
                printf("name: %s\nid:%d", a.name, a.id);
        }

        return 0;
}

This code compiles correctly but occurs Segmentation fault (core dumped) error.

`scanf("%d", tmp_id);` ==> `scanf("%d", &tmp_id);` or, even better, `if (scanf("%d", &tmp_id) != 1) exit(EXIT_FAILURE);` or even best, `char line[1000]; if (!fgets(line, sizeof line, stdin) exit(EXIT_FAILURE); if (sscanf(line, "%d", &tmp_id) != 1) exit(EXIT_FAILURE);` — pmg, Aug 11 '20 at 13:22
... and that has nothing in particular to do with structures. — John Bollinger, Aug 11 '20 at 13:22
`scanf("%s"` ==> `scanf("%9s"` or better: use `fgets` instead of `scanf` — Support Ukraine, Aug 11 '20 at 13:22
There's no need for `tmp_name` or `tmp_id`. `scanf("%d", &tmp_id);` ==> `scanf("%d",&(a.id));` — Andrew Henle, Aug 11 '20 at 13:24
We already have some thousands of "missing & in scanf" questions and none of them are particularly interesting to keep. Please just use the linked canonical dupe to close this FAQ. — Lundin, Aug 11 '20 at 13:39

Support Ukraine · Accepted Answer · 2020-08-11T16:22:01.497

You write:

This code compiles correctly ...

In that case I'll strongly recommend that you increase the warning level of your compiler.

For instance "gcc -Wall code.c" may give you a warning like:

In function 'main':
warning: format '%d' expects argument of type 'int *', but argument 2 has type 'int' [-Wformat=]
   15 |                 scanf("%d", tmp_id);
      |                        ~^   ~~~~~~
      |                         |   |
      |                         |   int
      |                         int *

which tell you everything, i.e. you pass an int to a function expecting an int pointer

So just pass an int-pointer like:

scanf("%d", &tmp_id);
            ^
            Take address of tmp_id so that you have a pointer

BTW: for gcc you can use -Werror so that all warnings is treated as errors

score 1 · Answer 2 · answered Aug 11 '20 at 13:24

1

You can directly assign value to struct. You don't need another variable.

struct Book a
scanf("%s", a.name);
scanf("%d", &a.id);
printf("name: %s\nid:%d", a.name, a.id);

answered Aug 11 '20 at 13:24

John Park

1,644
1
12
17

The `scanf("%s", a.name)` offers no protections against a buffer overflow vulnerability, and is no different than the `gets` function, in this respect. `scanf("%9s", a.name)` must have been used instead. – M. Nejat Aydin Aug 11 '20 at 13:32

score 1 · Answer 3 · answered Aug 11 '20 at 13:35

1

scanf("%d", tmp_id); has to be scanf("%d", &tmp_id);

answered Aug 11 '20 at 13:35

Michael Scholten

11
1

eggsu · Answer 4 · 2020-08-13T10:30:22.693

1

You need to pass in by reference when using scanf.

        scanf("%s", tmp_name);
        scanf("%d", &tmp_id);

edited Aug 13 '20 at 10:30

answered Aug 11 '20 at 13:42

eggsu

11
2

I think for `char` types there is no need to `&`. – Amir reza Riahi Aug 11 '20 at 13:44
1

@AmirrezaRiahi I agree, this way would make the beginners a wrong understanding of how C array really works. But I think it's not the "`char` types", it's the array. – user26742873 Aug 11 '20 at 14:18

How give value to structures in c?

4 Answers4