C ++，const_cast怀疑。如果我指向const变量并更改其值，为什么它保持不变？ [英] C++, const_cast doubt. If I point to const variable and change its value, why it remains the same?

问题描述

如果你有一个变量指向另一个变量，是不是应该更改它的值正弦更换值？

If you have a variable pointing at another, isn''t supposed to change its value sine its replacing the value?

#include <iostream>

using namespace std;

void print (char * str)
{
    *str = 'd';
  cout << *str << endl;
}

int main(void)
{
    const char c = 's';

    print(const_cast<char*>(&c));
    cout << c << endl;
    cin.get();

    return 0;
}

输出：



d

s

Output:

d
s

推荐答案

含义 const 已更改：



在C ++ 98中： const 意味着逻辑上是const，但是从C ++ 11开始 const 现在是线程安全的 - 意味着它必须是按位const。所以编译器可以自由重写：

The meaning const has been changed:

In C++98: const meant logically const, but as of C++11 const is now thread safe - meaning that it has to be bitwise const. So the compiler is free to rewrite:

cout << c << endl;

as：

as:

cout << ''s'' << endl;

按位常量意味着对象或变量中的每一位都是永久性的。



最好的问候

Espen Harlinn

Bitwise const means that every bit in the object or variable is permanent.

Best regards
Espen Harlinn

根据文档 [ ^ ]：

According to the documentation[^]:

Depending on the type of the referenced object, a write operation through the resulting pointer, reference, or pointer to data member might produce undefined behavior.

因为 c 是一个 const char 变量编译器可能正在将值保存在CPU寄存器中，而不是期望它更改，因此它永远不会从内存中重新读取它。结果是你看到的行为。



试试这个：

Because c is a const char variable the compiler is probably saving the value in a CPU register, not expecting it to change so it never actual re-reads it from memory. The result is the behaviour you are seeing.

Try this:

#include <iostream>

using namespace std;
 
void print (char * str)
{
    *str = 'd';
  cout << *str << endl;
}
 
int main(void)
{
    const char c = 's';
 
    print(const_cast<char*>(&c));
    cout << c << endl;
    cout << *(const_cast<volatile char *>(&c)) << endl;
    cout << c << endl;

    cin.get();
 
    return 0;
}

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