Struct node

[p595285902]

strcut Node * curPtr;
strcut Node  relation[10];

curPtr= (strcut Node * )relation[0];

while(curPtr->nextPtr != NULL)//find the last node of the list
                  {
                    printf("error check 1\n");
                    fflush(stdout);
                    curPtr =(struct Node)curPtr->nextPtr;
                    if(curPtr ==NULL)
                    {
                      printf("2");}
                  }

this is part of my code, the output is
error check1
error check1
Segmentation Fault (core dumped)

I ran it with gdb, it says that the line
[while(curPtr->nextPtr != NULL)//find the last node of the list]
is wrong. Need help!!

[quzah]

curPtr= (strcut Node * )relation[0];

Q. Why are you typcasting that?
A. Because my compiler complained that relation[0] wasn't a pointer.

That's because it wasn't a pointer! You shouldn't need to typecast assigning an address to a pointer of the same type.

curPtr = & relation[ X ];
curPtr = relation + X;

Pick one.


Quzah.

[p595285902]

I tried both
curPtr= & relation[x];
and
curPtr=relation + x;

but it gaves me
main.c:75: warning: assignment from incompatible pointer type

[quzah]

No it doesn't.

int main( void )
{
    struct foo { int x; } bar[2], *p = & bar[0];
    p = bar + 1;
    return 0;
}

Quzah.

[p595285902]

curPtr = & relation[i];//set curPtr to the array
                while(curPtr->nextPtr != NULL)//find the last node of the list
                  {
                    curPtr = curPtr->nextPtr;
                  }

this is part of mine;

oh i see whats wrong now, I creat struct node* relation[x];
thanks man

[p595285902]

struct node{char name[10];} relation[x];

char * str;
str= relation[x].name;

Is this the right way of getting name? When I try to use relation[x]->name it gives error.

[quzah]

It still never enters your mind that including the error you get is a good idea, does it?

int main( void )
{
    struct foo { char name[10]; } bar[2];
    char *p = bar[0].name;
    p = bar[1].name;
    return 0;
}

That's fine. So what are you really doing and what is the error you are getting?


Quzah.

[p595285902]

the error is:

invalid type argument of `->'

[quzah]

You don't have any -> operators in the sample you posted here. You can just post your actual segment of code that has the error in it, instead of typing it in.


Quzah.

[p595285902]

char *relname;
struct Tablenode {
  char * name;
  struct Tablenode *nextPtr;
} relation[x];

relname = relation[x]->name;//set relname

and the error is
invalid type argument of `->'

[quzah]

So stop using -> there. Did I use -> in my example? Did you use -> in the example where I said it was fine? No. So why are you using it here, especially when it just told you that it was wrong to do that.


Quzah.

[CommonTater]

char *relname;
struct Tablenode {
  char * name;
  struct Tablenode *nextPtr;
} relation[x];

relname = relation[x]->name;//set relname

and the error is
invalid type argument of `->'

Two reasons ...
1) C does not assign strings across the = sign.
2) It will assign a pointer, which you went ahead and dereferenced.

[quzah]

Two reasons ...
1) C does not assign strings across the = sign.
2) It will assign a pointer, which you went ahead and dereferenced.

1. There aren't actually any strings being assigned. He's assigning to a character pointer, which you can do in every case:

char foo[X];
char *bar = "";
char *baz;

baz = foo;
baz = bar;

Both possibilities here are perfectly legal.

2. He didn't dereference anything, because he wasn't working with a pointer to a structure, when he used the [x] notation. He could only use the arrow, if he did this:

struct foo { char name[X]; } bar[2];
char *baz;

baz = (bar + 1)->name;

That would have been fine. It didn't have anything to do with dereferencing name.


Quzah.