C/C++ 更改 const 的值

2022-01-23 00:00:00 constants c++

我有一篇文章，但我把它弄丢了.它展示并描述了一些人们应该小心的 C/C++ 技巧.其中一个对我很感兴趣，但现在我正在尝试复制它，我无法将它编译.

这个概念是可以在 C/C++ 中意外更改 const 的值

原来是这样的:

const int a = 3;//我保证我不会改变常量 int *ptr_to_a = &a;//我仍然保证我不会改变诠释*ptr；ptr = ptr_to_a;(*ptr) = 5;//我是个骗子;a 现在是 5

我想把这个给朋友看，但现在我错过了一步.有谁知道开始编译和工作缺少什么?

ATM 我得到从 'const int*' 到 'int*' 的无效转换，但是当我阅读这篇文章时，我尝试过，效果很好.

解决方案

你需要抛弃 constness:

linux ~ $ cat constTest.c#include <stdio.h>无效 modA(int *x){*x = 7;}诠释主要(无效){常量 int a = 3;//我保证我不会改变诠释*ptr；ptr = (int*)( &a );printf("A=%d
", a);*ptr = 5;//我是个骗子，a 现在是 5printf("A=%d
", a);*((int*)(&a)) = 6;printf("A=%d
", a);modA( (int*)( &a ));printf("A=%d
", a);返回0；}linux ~ $ gcc constTest.c -o constTestlinux ~ $ ./constTestA=3A=5A=6A=7linux ~ $ g++ constTest.c -o constTestlinux ~ $ ./constTestA=3A=3A=3A=3

通用答案在 g++ 4.1.2 中也不起作用

linux ~ $ cat constTest2.cpp#include <iostream>使用命名空间标准；诠释主要(无效){常量 int a = 3;//我保证我不会改变诠释*ptr；ptr = const_cast<int*>( &a );cout <

顺便说一句.. 永远不推荐这样做...我发现 g++ 不允许这种情况发生.. 所以这可能是您遇到的问题.

I had an article, but I lost it. It showed and described a couple of C/C++ tricks that people should be careful. One of them interested me but now that I am trying to replicate it I'm not being able to put it to compile.



The concept was that it is possible to change by accident the value of a const in C/C++

It was something like this:
const int a = 3;          // I promise I won't change a
const int *ptr_to_a = &a; // I still promise I won't change a
int *ptr;
ptr = ptr_to_a;

(*ptr) = 5;               // I'm a liar; a is now 5
I wanted to show this to a friend but now I'm missing a step.  Does anyone know what's missing for it to start compiling and working?

ATM I'm getting invalid conversion from 'const int*' to 'int*' but when I read the article I tried and it worked great.
 解决方案 you need to cast away the constness:
linux ~ $ cat constTest.c
#include <stdio.h>


void modA( int *x )
{
        *x = 7;
}


int main( void )
{

        const int a = 3; // I promisse i won't change a
        int *ptr;
        ptr = (int*)( &a );

        printf( "A=%d
", a );
        *ptr = 5; // I'm a liar, a is now 5
        printf( "A=%d
", a );

        *((int*)(&a)) = 6;
        printf( "A=%d
", a );

        modA( (int*)( &a ));
        printf( "A=%d
", a );

        return 0;
}
linux ~ $ gcc constTest.c -o constTest
linux ~ $ ./constTest
A=3
A=5
A=6
A=7
linux ~ $ g++ constTest.c -o constTest
linux ~ $ ./constTest
A=3
A=3
A=3
A=3
also the common answer doesn't work in g++ 4.1.2
linux ~ $ cat constTest2.cpp
#include <iostream>
using namespace std;
int main( void )
{
        const int a = 3; // I promisse i won't change a
        int *ptr;
        ptr = const_cast<int*>( &a );

        cout << "A=" << a << endl;
        *ptr = 5; // I'm a liar, a is now 5
        cout << "A=" << a << endl;

        return 0;
}
linux ~ $ g++ constTest2.cpp -o constTest2
linux ~ $ ./constTest2
A=3
A=3
linux ~ $
btw.. this is never recommended... I found that g++ doesn't allow this to happen.. so that may be the issue you are experiencing.


	
		相关文章