用 C 中的指针更改常量变量

Const variable changed with pointer in C(用 C 中的指针更改常量变量)

问题描述

变量 i 被声明为 const 但我仍然可以使用指向它的内存位置的指针来更改该值.怎么可能?

The variable i is declared const but still I am able to change the value with a pointer to the memory location to it. How is it possible?

int main()
{

    const int i = 11;
    int *ip = &i;
    *ip=100;
    printf("%d
",*ip);
    printf("%d
",i);
}

当我编译时，我收到此警告:

When I compile, I get this warning :

test.c: In function ‘main’:
test.c:11: warning: initialization discards qualifiers from pointer target type

输出是这个

100
100

推荐答案

const 不是对编译器的请求，以使其无法更改该变量.相反，这是对编译器的承诺，您不会这样做.如果你违背了你的承诺，你的程序就可以做任何事情，包括崩溃.

The const is not a request to the compiler to make it impossible to change that variable. Rather, it is a promise to the compiler that you won't. If you break your promise, your program is allowed to do anything at all, including crash.

例如，如果我使用带有 -O2 优化级别的 gcc 编译您的示例代码，输出为:

For example, if I compile your example code using gcc with the -O2 optimisation level, the output is:

100
11

编译器允许在只读内存中放置一个const限定的变量，但它没有(除了其他任何东西，有些环境没有实现任何这样的东西).特别是，将自动(本地")变量放在只读内存中几乎总是不切实际的.

The compiler is allowed to place a const qualified variable in read-only memory, but it doesn't have to (apart from anything else, some environments don't implement any such thing). In particular, it is almost always impractical for automatic ("local") variables to be placed in read-only memory.

如果将i的声明改为:

static const int i = 11;

那么你很可能会发现程序现在在运行时崩溃了.

then you may well find that the program now crashes at runtime.

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