What is the correct way of using calloc for x86 ISA

Question

I was shown many ways of using calloc and malloc: with casting, without, etc So here I have two option how I can use calloc. I am curious which one is right for x86 isa.

If I have the following:

typedef struct node{
    int numOfOccur; 
    int numOfSuperWords;
    struct node *children;
}NodePtr;

NodePtr* temp = &root;

How would be correct to allocate memory using calloc.

Option 1

temp -> children[currChar].children = (NodePtr *)calloc(27, sizeof(struct node));

Option 2

temp -> children[currChar].children = calloc(27, sizeof(children[currChar].children));

then why in old books we need all these casting.. but today people use it without casting and other stuff. Some things changed — YohanRoth, Oct 04 '14 at 18:35
The language might have changed. That's nothing to do with the ISA. — Oliver Charlesworth, Oct 04 '14 at 18:36
[In C you should not cast `malloc` and related functions](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). In fact, in C you should not cast any functions returning `void *`. It has nothing to do with the CPU instruction set. — Some programmer dude, Oct 04 '14 at 18:36
These "old" books are either bad (and therefore they are not edited anymore) or describe C++, which is another language. — Jens Gustedt, Oct 04 '14 at 18:39
Also your option 2 is wrong, because the expression for the `sizeof` operation is invalid, try it. — Jens Gustedt, Oct 04 '14 at 18:40

score 0 · Answer 1 · edited May 23 '17 at 12:20

0

There are several correct ways to write this. I would prefer the following:

temp->children[currChar].children = calloc(27, sizeof(struct node));

(i.e. use the type name instead of the long expression, and no cast.)

Apart from that, I am not terribly keen on the magic number (27).

edited May 23 '17 at 12:20

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answered Oct 04 '14 at 18:38

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What is the correct way of using calloc for x86 ISA

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