同一地址的变量如何产生两个不同的值?

How is a variable at the same address producing 2 different values?(同一地址的变量如何产生两个不同的值?)

问题描述

考虑一下:

#include <iostream>
using namespace std;

int main(void)
{
    const int a1 = 40;
    const int* b1 = &a1;
    char* c1 = (char *)(b1);
    *c1 = 'A';
    int *t = (int*)c1;


    cout << a1 << " " << *t << endl;
    cout << &a1 << " " << t << endl; 

    return 0;
}

这个输出是:

40 65 
0xbfacbe8c 0xbfacbe8c

除非编译器进行优化，否则这对我来说几乎是不可能的.如何 ?

This almost seems impossible to me unless compiler is making optimizations. How ?

推荐答案

This is 未定义行为，您正在修改一个 const 变量，因此您可以对结果没有任何期望.我们可以通过转到 C++ 标准草案部分 7.1.6.1 The cv-qualifiers 段落 4 来看到这一点，它说:

This is undefined behavior, you are modifying a const variable so you can have no expectation as to the results. We can see this by going to the draft C++ standard section 7.1.6.1 The cv-qualifiers paragraph 4 which says:

[...]在其生命周期 (3.8) 期间修改 const 对象的任何尝试都会导致未定义的行为.

[...]any attempt to modify a const object during its lifetime (3.8) results in undefined behavior.

甚至提供了一个例子:

const int* ciq = new const int (3); // initialized as required
int* iq = const_cast<int*>(ciq); // cast required
*iq = 4; // undefined: modifies a const object

在1.3.24部分的未定义行为的标准定义中，给出了以下可能的行为:

In the standard definition of undefined behaviour in section 1.3.24, gives the following possible behaviors:

[...] 允许的未定义行为的范围从完全忽略情况并产生不可预测的结果，到在翻译或程序执行期间以环境特征的记录方式行事(有或没有发布诊断消息)，以终止转换或执行(发布诊断消息).[...]

[...] Permissible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message). [...]

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