Code is as follows:
/* set.h */
struct setElement{
char *element;
setElement *next;
};
typedef struct setElement *Set; //Set is now the equivalent of setElement*
Set a;
setInit(&a);
/* setInit function declaration @ setInit.c */
int setInit(Set *a){
(*a)->element = "asdf"; //results in a seg fault
}
Trying to malloc 'a' works, but if I try to access any member within the set 'a' doesn't work. I understand I'm passing a reference of the set from the main() function to setInit, so I believe the pointer contained within setInit is addressing the memory allocated by 'Set a' in the main() function, so a malloc wouldn't be required...
Iunno. Help is appreciated :)
The problem is that you have not allocated the setElement
you are trying to assign to. In the main part of the code you are creating a Set
, which is just a pointer to a setElement
. This pointer is never set to point to anything sensible. Ie you need something like
Set a = malloc(sizeof(setElement));
Alas, it is unclear where exactly your variables are defined. I assume your main.c
is something like
#include "set.h"
Set a;
int main()
{
setInit(&a);
}
If so, your a, which is a pointer by itself, should point to somewhere.
If your framework wants malloc()
ed data, you should do
int main()
{
a = malloc(sizeof(*a)); // *a is a struct setElement now, with 2 pointer-sized members.
setInit(&a); // Now seInit should be able to operate on the struct as wanted.
}
As @amaurea has mentioned, you'll need to make use of malloc() for your setElement structure. In addition to this, you need to do the same for the setElement struct's element
member. A char*
is merely a pointer to a char or char array and will not implicitly allocate anything.
int setInit(Set *a){
(*a)->element = "asdf"; //results in a seg fault
}
Could be re-written
int setInit(Set *a){
(*a)->element = malloc(sizeof("asdf"));
strcpy((*a)->element,"asdf");
}
Which the above could be rewritten to take a second parameter of the actual element
contents.
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