在c或c ++中用于树创建的结构对象变量在点运算符和箭头运算符之间的区别 [英] difference between the dot operator and arrow operator by structure object variable for tree creation in c or c++

问题描述

我必须澄清一个疑问，即在c和c ++中也具有相同的概念.

I have to clear one doubt which has same concept in c and c++ as well.

假设我有一个这样的结构:

Suppose i have a struct like this:

struct Huffman
{
    int value;
    unsigned char sym;                 /* symbol */
    struct Huffman *left,*right;    /* left and right subtrees */
};

typedef struct Huffman Node;
Node * tree;

现在我使用树变量创建树.然后同时使用点运算符和箭头运算符.像这样.

and now i create tree using tree variable. Then using dot operator and arrow operator both. like this.

Arrorw运算符:

Arrorw operator:

 for (i = 0; i < data_size; i++) 
    {
         // the problem is here this tree pointer don't store the values for all alphabets, it just remembers the last executed alphabet after this for loop.
        tree -> left = NULL;
        tree  ->right = NULL;
        tree -> symbol = storesym[i];
        tree  -> freq = storefreq[i];
        tree -> flag = 0;
        tree -> next = i + 1;
        cout<<"check1 : "<<tree -> symbol<<endl;
    } 

点运算符:

for (i = 0; i < data_size; i++) 
{
    tree[i].symbol = storesym[i];
    tree[i].freq = storefreq[i];
    tree[i].flag = 0;
    tree[i].left = tree[i].right = tree[i].value = NULL;
    tree[i].next = i + 1;
}

现在我不明白(1)两者之间有什么区别?(2)如何在内存中分配它们?

推荐答案

.运算符期望其操作数是 struct ... 或联盟... .-> 运算符期望其操作数是指向 struct ... 的指针"或指向 union ... ".

The . operator expects its operand to be an expression of type struct ... or union .... The -> operator expects its operand to be an expression of type "pointer to struct ..." or "pointer to union ...".

表达式 tree 的类型为指向 struct Huffman 的指针"，因此您可以使用-> 运算符访问成员.

The expression tree has type "pointer to struct Huffman", so you use the -> operator to access a member.

表达式 tree [i] 的类型为" struct Huffman "；下标运算符隐式取消引用指针(请记住，将 a [i] 评估为 *(a + i))，因此您可以使用.操作员访问成员.

The expression tree[i] has type "struct Huffman"; the subscript operator implicitly dereferences the pointer (remember that a[i] is evaluated as *(a + i)), so you use the . operator to access a member.

基本上， a-> b 是(* a).b 的可读性更高的等效项.

Basically, a->b is the somewhat more readable equivalent of (*a).b.

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