C结构的指针初始化 | C结构的指针初始化

本文介绍了C结构的指针初始化的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

当包含结构指针数组的结构体实例化，
我保证，在结构数组成员的所有指针将被设置为 NULL ？

When a struct that contains an array of struct pointers is instantiated,am I guaranteed that all pointers in the struct array member will be set to NULL?

下面是一个例子结构：

typedef struct mmNode {
  int val;
  int board[2][NUM_PITS+1];
  int side;
  struct mmNode* children[NUM_PITS+1];
} mmNode;

IE：如果我创建mmNode结构的实例，将 mmNode.children 的元素总是被设置为 NULL ？

推荐答案

这取决于你如何初始化你的结构。

It depends how you initialise your struct.

mmNode a;                              // Everything default-initialized

void foo()
{
    static mmNode b;                   // Everything default-initialized

    mmNode  c;                         // Nothing initialized
    mmNode  d = { 0 };                 // Everything default-initialized
    mmNode *p = malloc(sizeof(*p));    // Nothing initialized
    mmNode *q = calloc(1, sizeof(*q)); // Everything zero-initialized
}

没有初始化意味着所有成员都将只是随机的垃圾值。默认初始化意味着所有成员将被初始化为0，这对于指针成员将等同于 NULL 。零初始化意味着一切都将被设置，位为0。这只能在平台上，其中 NULL 重新按位0 psented $ P $工作。

"Nothing initialized" means that all members will just have random junk values. "Default-initialized" means that all members will be initialized to 0, which for pointer members will be equivalent to NULL. "Zero-initialized" means that everything will be set, bitwise, to 0. This will only work on platforms where NULL is represented with bitwise 0.

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