是否可以将 C 指针初始化为 NULL? [英] Is it possible to initialize a C pointer to NULL?

问题描述

我一直在写类似的东西

char *x=NULL;

假设

 char *x=2;

将创建一个指向地址 2 的 char 指针.

would create a char pointer to address 2.

但是，在 GNU C 编程教程中 它说 int *my_int_ptr = 2; 将整数值 2 存储到分配时 my_int_ptr 中的任何随机地址.

But, in The GNU C Programming Tutorial it says that int *my_int_ptr = 2; stores the integer value 2 to whatever random address is in my_int_ptr when it is allocated.

这似乎意味着我自己的 char *x=NULL 正在分配 NULL 转换为 char 的任何值内存中的一些随机地址.

This would seem to imply that my own char *x=NULL is assigning whatever the value of NULL cast to a char is to some random address in memory.

虽然

#include <stdlib.h>
#include <stdio.h>

int main()
{
    char *x=NULL;

    if (x==NULL)
        printf("is NULL
");

    return EXIT_SUCCESS;
}

确实会打印

为空

当我编译和运行它时，我担心我依赖于未定义的行为，或者至少是未指定的行为，我应该写

when I compile and run it, I am concerned that I am relying on undefined behavior, or at least under-specified behavior, and that I should write

char *x;
x=NULL;

改为.

推荐答案

是否可以将 C 指针初始化为 NULL?

TL;DR是的，非常.

实际声明指南看起来像

另一方面，如果你只使用单个初始赋值，int *my_int_ptr = 2;，程序将尝试填充由 my_int_ptr 的值为 2.由于 my_int_ptr 被垃圾填充，它可以是任何地址.[...]

On the other hand, if you use just the single initial assignment, int *my_int_ptr = 2;, the program will try to fill the contents of the memory location pointed to by my_int_ptr with the value 2. Since my_int_ptr is filled with garbage, it can be any address. [...]

嗯，他们是错了，你是对的.

Well, they are wrong, you are right.

对于语句，(暂时忽略指向整数转换的指针是实现定义的行为这一事实)

int * my_int_ptr = 2;

my_int_ptr 是一个变量(类型为指向 int 的指针)，它有自己的地址(类型:指向整数的指针的地址)，您正在存储一个2 的值转换为 那个 地址.

my_int_ptr is a variable (of type pointer to int), it has an address of its own (type: address of pointer to integer), you are storing a value of 2 into that address.

现在，my_int_ptr，作为一个指针类型，我们可以说，它指向type"的值.在 指向 my_int_ptr 中保存的值的内存位置.因此，您实际上是在为指针变量分配 of 的值，而不是指针指向的内存位置的值.

Now, my_int_ptr, being a pointer type, we can say, it points to the value of "type" at the memory location pointed by the value held in my_int_ptr. So, you are essentially assigning the value of the pointer variable, not the value of the memory location pointed to by the pointer.

所以，总结一下

 char *x=NULL;

将指针变量x初始化为NULL，而不是指针指向的内存地址处的值.

initializes the pointer variable x to NULL, not the value at the memory address pointed to by the pointer.

这与

 char *x;
 x = NULL;    

---

扩展:

现在，严格遵守，像这样的陈述

---

Expansion:

Now, being strictly conforming, a statement like

 int * my_int_ptr = 2;

是非法的，因为它涉及违反约束.说清楚，

is illegal, as it involves constraint violation. To be clear,

• my_int_ptr是指针变量，类型int *
• 一个整数常量，2 根据定义具有 int 类型.

• my_int_ptr is a pointer variable, type int *
• an integer constant, 2 has type int, by definition.

它们不是兼容的"；类型，所以这个初始化是无效的，因为它违反了简单赋值的规则，在第 6.5.16.1/P1 章中提到，在 Lundin 的回答中描述.

and they are not "compatible" types, so this initialization is invalid because it's violating the rules of simple assignment, mentioned in chapter §6.5.16.1/P1, described in Lundin's answer.

如果有人对初始化如何与简单的赋值约束相关联感兴趣，请引用 C11，第 6.7.9 章，P11

In case anyone's interested how initialization is linked to simple assignment constraints, quoting C11, chapter §6.7.9, P11

标量的初始值设定项应该是一个表达式，可选地用大括号括起来.这对象的初始值是表达式的初始值(转换后)；同一类型适用于简单赋值的约束和转换，采用标量的类型成为其声明类型的非限定版本.

The initializer for a scalar shall be a single expression, optionally enclosed in braces. The initial value of the object is that of the expression (after conversion); the same type constraints and conversions as for simple assignment apply, taking the type of the scalar to be the unqualified version of its declared type.

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