初始化二叉搜索树中的字符串数组

Question

my node contains an int and a string variable, and I tried using binary search tree. the code is as follow:

struct node{
int a;
string members[5];
};

int main(){
node * root = NULL;
root = (node*)malloc(sizeof(node));
root->members[0] = "aaaaa";
return 0;
}

of course, my code wasn't exactly like that, I made it short in main because I want to show just the problem. it gave me 'access violation writing location'. I tried using 'new node();' instead of malloc and that didn't happen. why is this exactly?

Answer 1

malloc() only allocates memory. It doesn't call the constructor of the object. You could call the constructor on allocated memory using, eg

void* mem = malloc(sizeof(node));
if (mem) {
    node* root = new(mem) node;
    // ...
}

When using new node instead of malloc(sizeof(node) the allocate memory also gets initialized. Using an uninitialized object is undefined behavior.

Answer 2

malloc only allocates raw storage. new allocates raw storage and initializes it to contain a specified type.

If you're allocating only POD types, that distinction is mostly one of wording, with little real difference in what happens.

If you're allocating something like an std::string that has a constructor, there's a world of difference though. If you use new , then your string variables have all been initialized so they're real (albeit, still empty) strings. When you just use malloc , they haven't been initialized at all--they're just chunks of uninitialized raw storage the right size to contain a string . When you try to use them as a string, a quick crash is about the best you can hope for.