How does struct Node *ptr = malloc(sizeof(*ptr)) work?

Question

I'm a C beginner, and I came across this code while trying to implement a linked list.

    struct Node *ptr = malloc(sizeof(*ptr));

The Node struct looks like this:

    struct Node {
        int data;
        struct Node *next;
    };

I'm trying to understand the first line. It seems as if malloc(sizeof(*ptr)) already knows the contents of ptr. What exactly is happening on the left side and is it happening before malloc is called?

Answer 1

It seems as if malloc(sizeof(*ptr)) already knows the contents of ptr.

Actually, it doesn't. The sizeof operator doesn't actually evaluate its operand (unless it's a variable length array), it just looks at its type. This means that ptr isn't actually dereferenced and is therefore a safe operation.

Answer 2

Only use paranthesis with sizeof if there's a type you'd like to know the size of. If you have an expression, like *ptr, it's enough to write:

struct Node *ptr = malloc(sizeof *ptr);         // <- no parenthesis

The expression *ptr dereferences the pointer so it becomes a struct Node and that's what the sizeof is returning the size for.

sizeofexpression - Returns the size, in bytes, of the object representation of the type of expression. No implicit conversions are applied to expression.

It's the same size you get if you do:

struct Node *ptr = malloc(sizeof(struct Node)); // <- parenthesis needed

but the first alternative is often preferable for clarity.