Trying to set a string in a struct

Question

struct list_el {
    unsigned int data;
    char name[16];
    char grade;
    struct list_el * next;
};

typedef struct list_el item;

int main(int argc, char **argv) {
    item *curr, *head;

    //first item
    curr->data = 3141592;
    strcpy(curr->name, "Carl");
    curr->grade = 'A';
    curr->next = head;
    head = curr;

Trying to figure out why this isn't working when I try to set name to "Carl". I'm getting "too few arguments to function 'strcpy'" even though I have 2 arguments in it (destination, source). When I add a 3rd argument (how many characters to copy?), I end up getting "assignment to expression with array type".

Post an MCVE and you haven't allocated memory for curr before using it. — Spikatrix
– Spikatrix, Commented May 1, 2015 at 5:37
Not sure why this would fail to compile but it's definitely UB - you're declaring a pointer to an item but have not initialized it to an actual instance. — MooseBoys
– MooseBoys, Commented May 1, 2015 at 5:40
I'm sorry, this is preventing eclipse from building my project for me... — CodingNinjaInTraining
– CodingNinjaInTraining, Commented May 1, 2015 at 5:40
it's working after allocating memory x.x Thanks for the help! Wonder why it didn't give me an error for the rest of it... — CodingNinjaInTraining
– CodingNinjaInTraining, Commented May 1, 2015 at 5:45

geek_sandeep · Accepted Answer · 2015-05-01 06:10:14Z

I am not getting your error "too few arguments to function 'strcpy' " but i can see that you have declare a item *curr but never initialized.So i tried with correcting this and updated your code as below and its not giving any error.

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct list_el {
    unsigned int data;
    char name[16];
    char grade;
    struct list_el * next;
};

typedef struct list_el item;

int main(int argc, char **argv) {
    item *curr, *head;
    curr = malloc(sizeof(struct list_el));
    //first item
    curr->data = 3141592;
    strcpy(curr->name, "Carl");
    curr->grade = 'A';
    curr->next = head;
    head = curr;
}

I think, it will work for you.

Idea: rather than curr = malloc(sizeof(struct list_el));, consider curr = malloc(sizeof *curr); Easier to code, less likely to be wrong and easy to maintain.

Spikatrix · Accepted Answer · 2015-05-01 06:00:11Z

1

You did not allocate memory for curr before using it. Two options:

Make curr point to an instance of item:
```
item i;
curr = &i;
```
Allocate memory dynamically for curr using malloc/calloc:
```
curr = malloc(sizeof(*curr));
/*OR*/
curr = calloc( 1, sizeof(*curr));
```
and later, free it after its use. It is also a good idea to check if malloc/calloc did not fail by checking its return value. It will return NULL on failure.

answered May 1, 2015 at 6:00

Spikatrix

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