malloc（sizeof（struct Node））与malloc（sizeof（nodeptr））的不同行为

Question

I was trying to allocate memory using malloc, I am not able to understand why I am getting a different result for these two malloc calls.

The line below gives me wrong result even though with gdb I see the data is getting the correct value assigned.

• nodeptr n = malloc(sizeof(nodeptr));

  Value head->data: '!'
  Value head->eq->data: ''

And when I do this get the correct result:

• nodeptr n = malloc(sizeof(struct Node));

  Value head->data: 'w'
  Value head->eq->data: 'X'

I followed this post, I think I am doing it correctly.
In both ways, while allocation I get the same amount of memory, just I see the different results in the end.

typedef struct Node
{
    struct Node *left, *right, *eq;
    char data;
    bool isEnd;
} *nodeptr;

nodeptr newNode(const char c) {
    nodeptr n = malloc(sizeof(nodeptr));
    // nodeptr n = malloc(sizeof(struct Node));
    n->data = c;
    n->left = NULL;
    n->right = NULL;
    n->left = NULL;
    n->isEnd = false;

    return n;
}

void insert(nodeptr *node, const char *str) {

    if (*node == NULL) {
        *node = newNode(*str);
    }
    nodeptr pCrawl = *node;

    if(pCrawl->data < *str) {
        insert(&pCrawl->right, str);
    } else if (pCrawl->data > *str) {
        insert(&pCrawl->left, str);
    } else {
        if(*(str+1)) {
            insert(&pCrawl->eq, str + 1);
        } else {
            pCrawl->isEnd = true;
        }
    }
}

int main(int argc, char const *argv[])
{

    const char* const strs[5]= {
        "w.",
    };
    nodeptr head = NULL;
    for(int i = 0; i<1; i++) {
        insert(&head, strs[i]);
    }
    return 0;
    printf("Value head->data: \'%c\'\n", head->data);
    printf("Value head->eq->data: \'%c\'\n", head->eq->data);
}

Answer 1

The two different versions aren't allocating the same amount of memory. sizeof(nodeptr) is the size of a pointer and sizeof(struct Node) is the size of your struct . These are not the same things and they're not the same size. On my computer these values are 8 and 32 .

You want to use:

nodeptr n = malloc(sizeof(struct Node));

or perhaps:

nodeptr n = malloc(sizeof(*n)); // size of the type that n points too

Answer 2

sizeof(nodeptr) == sizeof(struct Node*) != sizeof(struct Node) == sizeof(*nodeptr)

• sizeof(nodeptr) will always be the size of a pointer (like 8 bytes on a 64-bit CPU)
• sizeof(struct Node) refers to the struct contents
• sizeof(*nodeptr) is equivalent to sizeof(struct Node) with the extra dereference operator in there.

The reason it may appear to "work" (not segfault) is that malloc suballocates from a larger block of heap memory. However, the code is writing out-of-bounds of the requested allocation, which can eventually lead to heap corruption or segfaults at some point.