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

user_13451528 0
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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?

malloc pointers sizeof c
3 Answers

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.

user_75822470

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.

user_16871190

You need to allocate a memory for an object of the type struct Node declared like

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

So in the call of malloc you need to specify the size of the memory to be allocated for an object of this type. The function malloc returns a pointer to the allocated memory.

So you can write

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

On the other hand, the expression *ptr has the type struct Node. That is the declared pointer ptr has the pointer type struct Node * and dereferencing such a pointer like *ptr yields an expression of the type struct Node.

So you may rewrite the above record also like

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

because in this case sizeof( struct Node ) is equivalent to sizeof( *ptr ).

That is the compiler needs to know the type of the operand of the operator sizeof that to determine the size of an object of that type.

user_28772410
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