指针数组和C中的只读内存 [英] Pointers Arrays and Read Only Memory in C

问题描述

我正在学习C编程语言，并且在理解指针，数组和只读存储器之间的细微差别时遇到了困难.我正在使用以下示例:

I am learning the C programming language and am running into difficulties in understanding the minute differences between pointers, arrays and read only memory. I am working with the following example:

char *myCards = "JQK";

据我了解，这行代码完成的工作是，它在可通过char指针myCards访问的只读存储区中创建了一个以空终止的字符字符串.这意味着我能够编译以下内容，但由于字符串的不可变性而会收到错误:

From what I understand what this line of code accomplishes is it creates a null terminated string of characters in a read only section of memory that is accessible via the char pointer myCards. This means that I am able to compile the following but would receive an error due to the immutability of the string:

*myCards = 'A';

我正在尝试使用整数数组来实现类似的功能；但是，我完全不确定如何创建此数组并将其放置在内存的只读部分中.

I am trying to achieve a similar functionality using an array of ints; however, I am not at all certain on how to create this array and place it in a read only section of memory if at all possible.

我知道我可以使用const关键字并按如下方式创建一个数组:

I know that I can use the const keyword and create an array as follows:

const int myInts[] = {3, 6, 1, 2, 3, 8, 4, 1, 7, 2};

但是，在此数组初始化之后，我能够直接执行以下操作:

However, I am able to do the following, directly after this array initialization:

printf("First element of array: %i\n", myInts[0]);    
int *myIntsPtr = myInts;
*myIntsPtr = 15;
printf("First element of array: %i\n", myInts[0]);

我能够通过创建指向数组的第一个元素来更改该数组的第一个元素，这表明该数组从未被置于只读存储器中.

I was able to alter the first element of the array by creating a pointer to it, implying that this array was never placed into read only memory.

基本上，我一直试图找出如何声明此int数组，以便使其位于类似于"JQK"的只读内存中.任何见解都是有帮助的.谢谢.

Basically I am stuck on trying to figure out how I am able to declare this int array so that it is in read only memory similar to "JQK". Any insight is helpful. Thank you.

推荐答案

好，首先要了解这一点，重要的是要知道C中的const不必对只读内存做任何事情.对于C，没有诸如section之类的东西. const仅仅是一个合约，它表达的意图是某些东西确实是恒定的.这意味着编译器/链接器 可以将数据放置在只读区域中，因为程序员保证数据不会更改.但是，它没有.

Ok, first to understand this, it's important to know that const in C doesn't have to do anything with read-only memory. For C, there is no such thing as sections. const is merely a contract, it's expressing the intention that something is indeed constant. This means a compiler/linker can place data in a read-only section because the programmer assured it won't change. It doesn't have to, though.

第二，字符串文字会转换为char的常量数组，并隐式附加0.请在此处查看Peter Schneider的评论:它不是形式上 const(因此，当您使用非const指针时，编译器不会警告您)，但它应该

Second, a string literal translates to a constant array of chars with 0 implicitly appended. See Peter Schneider's comment here: it is not formally const (so the compiler won't warn you when you take a non-const pointer to it), but it should be.

与此相结合，我的系统上的以下代码段在Linux amd64上与gcc一起在我的系统上出现，因为gcc确实将数组放置在只读段中:

Combining this, the following code segfaults on my system with gcc on Linux amd64, because gcc indeed places the array in a read-only section:

#include <stdio.h>

const int myInts[] = {3, 6, 1, 2, 3, 8, 4, 1, 7, 2};

int main(void)
{
    printf("First element of array: %i\n", myInts[0]);    
    int *myIntsPtr = myInts;
    *myIntsPtr = *(myIntsPtr + 1);
    printf("First element of array: %i\n", myInts[0]);
    return 0;
}

请注意，在将非const指针指向const数组的行中，还会出现编译器警告.

Note there is also a compiler warning in the line where you take a non-const pointer to the const array.

顺便说一句，如果您使用gcc在函数中声明数组，则相同的代码将起作用，这是因为然后，数组本身是在堆栈上创建的.仍然会收到警告，代码仍然是错误的.这是有关如何在此处实现C的技术细节.与字符串文字的不同之处在于，它是一个匿名对象(char数组没有标识符)，并且在任何情况下都具有静态存储期限.

Btw, the same code will work if you declare the array inside your function with gcc, that's because then, the array itself is created on the stack. Still you get the warning, the code is still wrong. It's a technical detail of how C is implemented here. The difference to a string literal is that it is an anonymous object (the char array doesn't have an identifier) and has static storage duration in any case.

编辑来解释字符串文字的作用:以下代码是等效的:

edit to explain what a string literal does: The following codes are equivalent:

int main(void)
{
    const char *foo = "bar";
}

和

const char ihavenoname_1[] = {'b', 'a', 'r', 0};

int main(void)
{
    const char *foo = ihavenoname_1;
}

---

因此，简而言之，如果要gcc将数据放在只读节中，请使用静态存储持续时间(函数外部)将其声明为const.其他编译器的行为可能有所不同.

---

So, short story, if you want gcc to put data in a read-only section, declare it const with static storage duration (outside of a function). Other compilers might behave differently.

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