C / C ++改变const的值 [英] C/C++ changing the value of a const

问题描述

我有一篇文章，但我失去了。它显示和描述了几个C / C ++的技巧，人们应该小心。其中一个有兴趣我，但现在我想复制它，我不能把它编译。<​​/ p>

这个概念是可以改变意外的C / C ++中的 const 的值

这是这样的：

  const int a = 3; //我保证我不会改变
 const int * ptr_to_a =& a; //我仍然保证我不会改变一个
 int * ptr; 
 ptr = ptr_to_a; 
 
（* ptr）= 5; //我是骗子a现在是5 
  

我想向朋友展示这个，但现在我错过了一步。

ATM我得到从'const int *'到'int *'的无效转换。 / strong>但是当我阅读我尝试的文章，它工作得很好。

解决方案

你需要抛弃的constness： / p>

  linux〜$ cat constTest.c 
 #include< stdio.h> 
 
 
 void modA（int * x）
 {
 * x = 7; 
} 
 
 
 int main（void）
 {
 
 const int a = 3; //我promisse我不会改变一个
 int * ptr; 
 ptr =（int *）（& a）; 
 
 printf（A =％d\\\
，a）; 
 * ptr = 5; //我是骗子，现在是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 
 = 3 
 A = 3 
  

同样的答案在g ++ 4.1中不起作用。 2

  linux〜$ cat constTest2.cpp 
 #include< iostream> 
 using namespace std; 
 int main（void）
 {
 const int a = 3; //我promisse我不会改变一个
 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 ..这是不推荐的...我发现g ++不允许这种情况发生..因此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\n", a );
        *ptr = 5; // I'm a liar, a is now 5
        printf( "A=%d\n", a );

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

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

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