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

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编程教程表示int *my_int_ptr = 2;将整数值2存储到my_int_ptr中分配的随机地址中.

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.

这似乎暗示我自己的char *x=NULL正在将强制转换为char的NULL值分配给内存中的某个随机地址.

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.

而

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

int main()
{
    char *x=NULL;

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

    return EXIT_SUCCESS;
}

实际上可以打印

为NULL

is NULL

当我编译并运行它时，我担心我依赖未定义的行为，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;

相反.

推荐答案

是否可以将C指针初始化为NULL?

TL; DR 是的，非常.

根据指南的内容类似于

The actual claim made on the guide reads like

另一方面，如果仅使用单个初始分配int *my_int_ptr = 2;，则程序将尝试用值2填充my_int_ptr指向的内存位置的内容. >充满垃圾，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. [...]

好吧，他们 是错误的，你是对的.

Well, they are wrong, you are right.

对于该语句，(暂时忽略指向整数转换的指针是实现定义的行为的事实)

For the statement, (ignoring, for now, the fact that pointer to integer conversion is an implementation-defined behaviour)

int * my_int_ptr = 2;

my_int_ptr是一个变量(类型为int的指针)，它具有其自己的地址(类型:指向整数的指针的地址)，您要将2的值存储到中， 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是指针类型，可以说，它指向在由指向所指向的存储位置处的"type"的值在my_int_ptr中.因此，您实际上是在给指针变量 of 赋值，而不是分配指针所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，要得出结论

 char *x=NULL;

将指针变量x初始化为NULL，而不是指针所指向的内存地址的值.

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

这与相同

 char *x;
 x = NULL;    

---

扩展:

现在，严格遵循这样的声明

---

Expansion:

Now, being strictly conforming, a statement like

 int * my_int_ptr = 2;

是非法的，因为它涉及约束违反.要清楚，

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

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

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

并且它们不是兼容"类型，因此此初始化无效，因为它违反了伦丁的答案.

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.

如果有人对初始化与简单分配约束的联系方式感兴趣，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为单个表达式，可以选择用大括号括起来.这 对象的初始值是表达式的初始值(转换后)； 相同类型 适用于简单分配的约束和转换，采用标量的类型 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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