Thread: Puzzled

struct LLNode *initLLNode(char val){
struct LLNode *temp;
temp=(struct LLNode *)malloc(sizeof(struct LLNode));
temp->next=NULL;
temp->key=val;
return(temp);
}

i have these lines of code...the line which puzzles me is the line with (struct LLNode *) next to malloc
wat is its purpose there?

malloc returns a void pointer (pointer of no particular type but one that can be cast to become of any type) so if you casted it as struct LLNode *, then malloc will return that type.

correct me if im wrong

>correct me if im wrong
You're wrong. C allows implicit conversion from a void* to any pointer type (except function pointers). The cast is unnecessary and dangerous as it potentially hides the error of not including stdlib.h. The preferred way to call malloc in the C community is as follows:

Code:

temp = malloc(sizeof (*temp));

if (temp)
{
    /* Use the pointer safely now */
}
else
{
    /* Recover from a failed call to malloc */
}

i see, i don't get the part when u said it po tentially hides the error of not including stdlib.h, where does this stdlib.h suddenly get involved in here?

>where does this stdlib.h suddenly get involved in here?
malloc is declared in stdlib.h. If you fail to include the standard header in which a function is declared, any use of that function is undefined. In short, your program is very wrong and will fail to link if you're lucky. If you're not lucky then it will run, but with unpredictable behavior.

so the problem with using this kind of malloc assignment is dangerous?

but i still don't see how dangerous it can be by using this way, probably u could help me by elaborating further?