如果 char*s 是只读的，为什么我可以覆盖它们? [英] If char*s are read only, why can I overwrite them?

问题描述

我的课程告诉我 char*s 是静态的/只读的，所以我认为这意味着你在定义它们后无法编辑它们.但是当我跑步时:

My course taught me that char*s are static/read only so I thought that would mean you can't edit them after you have defined them. But when I run:

char* fruit = "banana";
printf("fruit is %s\n", fruit);
fruit = "apple";
printf("fruit is %s\n", fruit);

然后它编译得很好并给了我:

Then it compiles fine and gives me:

fruit is banana
fruit is apple

为什么?我是否误解了只读意味着什么?对不起，如果这很明显，但我是编码新手，我无法在网上找到答案.

Why? Have I misunderstood what it means to be read-only? Sorry if this is obvious but I'm new to coding and I can't find the answer online.

推荐答案

呈现的代码片段不会更改字符串文字本身.它只改变存储在指针 fruit 中的值.

The presented code snippet does not change the string literals themselves. It only changes the values stored in the pointer fruit.

你可以想象这些线条

char* fruit = "banana";
fruit = "apple";

以下方式

char unnamed_static_array_banana[] = { 'b', 'a', 'n', 'a', 'n', 'a', '\0' };
char *fruit = &unnamed_static_array_banana[0];
char unnamed_static_array_apple[]  = { 'a', 'p', 'p', 'l', 'e', '\0' };
fruit = &unnamed_static_array_apple[0];

这些语句不会更改与字符串文字对应的数组.

These statements do not change the arrays that correspond to the string literals.

另一方面，如果你会尝试写作

On the other hand if you will try to write

char* fruit = "banana";
printf("fruit is %s\n", fruit);
fruit[0] = 'h';
^^^^^^^^^^^^^^
printf("fruit is %s\n", fruit);

也就是说，如果您尝试使用指向它的指针(指向字符串文字的第一个字符)来更改字符串文字，那么程序将具有未定义的行为.

that is if you will try to change a string literal using a pointer that points to it (to the first character of the string literal) then the program will have undefined behavior.

来自 C 标准(6.4.5 字符串文字)

From the C Standard (6.4.5 String literals)

7 未指定这些数组是否不同，只要它们元素具有适当的值.如果程序试图修改这样的数组，行为未定义.

7 It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.

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