在C语言中，可以一个const变量通过指针进行修改？ [英] In C, can a const variable be modified via a pointer?

问题描述

我写了类似这样的一些东西在我的code

I wrote some thing similar to this in my code

const int x=1;
int *ptr;
ptr = &x;
*ptr = 2;

关于是否所有的编译器这项工作吗？为什么不，我们正在改变一个常数变量GCC编译器通知？

Does this work on all compilers? Why doesn't the GCC compiler notice that we are changing a constant variable?

推荐答案

常量其实并不意味着常量。东西是常量用C具有的在编译时确定的值;文字 42 就是一个例子。在常量关键字的真正含义是只读的。举个例子：

const actually doesn't mean "constant". Something that's "constant" in C has a value that's determined at compile time; a literal 42 is an example. The const keyword really means read-only. Consider, for example:

const int r = rand();

的研究没有确定，直到程序执行时间的价值，但常量关键字意味着你不允许修改研究它的初始化之后。

The value of r is not determined until program execution time, but the const keyword means that you're not permitted to modify r after it's been initialized.

在您的code：

const int x=1;
int *ptr;
ptr = &x;
*ptr = 2;

分配 PTR =安培; X; 是约束违反的，也就是说，一个符合标准的编译器需要抱怨;你不能合法地分配 const int的* （指向INT常数）值，以一个非const 为int * 目的。的如果的编译器生成一个可执行文件（它不需要做;它可能只是拒绝），则行为不是由C标准定义

the assignment ptr = &x; is a constraint violation, meaning that a conforming compiler is required to complain about it; you can't legally assign a const int* (pointer to const int) value to a non-const int* object. If the compiler generates an executable (which it needn't do; it could just reject it), then the behavior is not defined by the C standard.

例如，生成code实际上可能存储在值 2 X - 但随后稍后参考 X 可能产生的价值 1 ，因为编译器的知道的是 X 不能被初始化后修改。而且它知道，因为你告诉它的话，通过定义 X 为常量。如果你骗编译器，后果可以任意坏的。

For example, the generated code might actually store the value 2 in x -- but then a later reference to x might yield the value 1, because the compiler knows that x can't have been modified after its initialization. And it knows that because you told it so, by defining x as const. If you lie to the compiler, the consequences can be arbitrarily bad.

其实，这可能发生的最糟糕的事情是，当你期待它的表现的程序;这意味着你有一个非常难以察觉的错误。 （但你应该得到的诊断将是一个大的线索。）

Actually, the worst thing that can happen is that the program behaves as you expect it to; that means you have a bug that's very difficult to detect. (But the diagnostic you should have gotten will have been a large clue.)

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