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C ++ what is the difference between usage. or & # 8594; for linked list

Suppose I have a structure.

struct node {
 string info;
 node link*;

}

What's the difference between

node k;
node b;

k.info = "string";
b.info = "string";
k.link = &b;

and

node *k;
node *b;
k = new node;
b = new node;    
k->info = "string";
b->info = "string";
k->link = b;

in terms of memory allocation? Are both examples correct and create an appropriate linked list? Added: Most books use a second example, why? Is there a downside to using the first example?

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9 answers

Yes. both are technically correct.

In the first example, the memory is on the stack, the second is on the heap.

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node k, *m, *n;

m = &k;
n = new node;

k.info  = "Hello";   // k is a node type object so use directly . operator
m->info = "Hi";      // m is a pointer to an object to type node, so use -> operator
n->info = "Man";     // n is a pointer to an object to type node, so use -> operator
*(m).info = "This";  // *(m) refers to an object itself, we use . operator on it
*(n).info = "Is a test"; // *(n) refers to an object itself, we use . opeartor on it

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→ . : , List, getRoot(), Nodes. getNext(), node .

*List myList;

Suppose we know that myList is a 5-node list. The list and 5 nodes are located on the heap (dynamically allocated). Now we want to access the last node.

myList->getRoot()->getNext()->getNext()->getNext()->getRoot();
(*(*(*(*(*myList).geRoot() ).getNext()).getNext()).getNext()).getNext();

These two sentences are equivalent and will return the fifth and last node. For the convenience of the programmer, use the first one.

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