Implement an array of linked lists

Question

I would like to implement an array of linked lists (in order to implement a hash table).

I have a problem at the very beginning... I defined a linked list structure, and I initialized the array of linked lists like an array of pointer to a list : liste * Oblist[65003]; (Do i need to initialize it like this : liste * Oblist[65003] = {NULL};?)

And I want to get the list at the index 12121 (which is NULL for the moment), so I did

liste L = *Oblist[12121] ;

The program compiles without errors but when I run it I get a Segmentation fault (core dumped)

Here is the whole code :

typedef struct Doublet {char * car ; struct Doublet * cdr ;}* liste ;

liste * Oblist[65003];

int main(void){

    liste L = *Oblist[32123] ;

    return 0;
}

Thank you in advance ;)

You seem to confuse yourself with the typedefed pointer. You have an array of pointers to pointers of type struct Doublet. — mafso
– mafso, Commented Jul 29, 2014 at 22:32
Yep, probably, but I already used linked lists, several times, without problem. — Simon
– Simon, Commented Jul 29, 2014 at 22:33

Rudy Velthuis · Accepted Answer · 2014-07-29 22:37:33Z

1

You have a few problems. liste is typedef'ed as a pointer to a struct Doublet, so it should be

liste Oblist[65003];

...

liste L = Oblist[32123];

Or, alternatively:

typedef struct Doublet { ... etc. } liste;

...

liste *Oblist[65003];

...

liste *list = Oblist[32123];

answered Jul 29, 2014 at 22:37

Rudy Velthuis

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3 Comments

Simon Over a year ago

Oh, seems like I'm a bit tired ;) Thanks for your answer!

Simon Over a year ago

Just one little question more : if I define my array of lists like this : liste Oblist[65003]; and I make it a global variable, how should I define in the other files this variable : extern + ... ? Thanks in advance ;)

Simon Over a year ago

I suppose that it is an other stupid question and the answer is extern list Oblist[60003], no?

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