g++ 是否满足 std::string C++11 要求

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问题描述

考虑以下示例:

int main()
{
    string x = "hello";
    //copy constructor has been called here.
    string y(x);
    //c_str return const char*, but this usage is quite popular.
    char* temp = (char*)y.c_str();

    temp[0] = 'p';

    cout << "x = " << x << endl;
    cout << "y = " << y << endl;

    cin >> x;
    return 0;
}

在 Visual Studio 编译器和 g++ 上运行它.当我这样做时，我得到了两个不同的结果.
在 g++ 中:

Run it on visual studio compiler and on g++. When I did so, I got two different results.
in g++:

x = pello  
y = pello

在视觉工作室 2010 中:

In visual studio 2010:

x = hello  
y = pello

产生差异的原因很可能是 g++ std::string 实现使用了 COW(写入时复制)技术，而 Visual Studio 没有.

The reason for the diff is most likely that g++ std::string implementation uses COW (copy on write) techniques and visual studio does not.

现在 C++ 标准(第 616 页表 64)说明了字符串复制构造函数

Now the C++ standard (page 616 table 64) states with regards to string copy constructor

basic_string(const basic_string& str):

效果:
data() 应该指向数组的已分配副本的第一个元素，该数组的第一个元素由 str.data() 指向"

effects:
data() should "points at the ﬁrst element of an allocated copy of the array whose ﬁrst element is pointed at by str.data()"

意思是 COW 是不允许的(至少在我的理解中).
怎么可能?
g++ 是否满足 std::string C++11 要求?

Meaning COW is not allowed (at least to my understanding).
How can that be?
Does g++ meets std::string C++11 requirements?

在 C++11 之前，这并没有造成大问题，因为 c_str 没有返回指向字符串对象保存的实际数据的指针，因此更改它并不重要.但是在更改之后，这种 COW + 返回实际指针的组合可以并且破坏旧的应用程序(由于编码错误而值得它的应用程序，但尽管如此).

Before C++11 this did not pose a big problem since c_str didn't return a pointer to the actual data the string object holds, so changing it didn't matter. But after the change this combination of COW + returning the actual pointer can and breaks old applications (applications that deserve it for bad coding but nevertheless).

你同意我的观点吗?如果是，可以做些什么吗?有没有人知道如何在一个非常大的旧代码环境中处理它(抓住这个的发条规则会很好).

Do you agree with me? If yes, can something be done? Does anyone have an idea about how to go at it in a very big old code environments (a clockwork rule to catch this would be nice).

请注意，即使不强制转换 const，也可能会通过调用 c_str、保存指针然后调用非 const 方法(这将导致写入)导致指针无效.
另一个没有抛弃 constness 的例子:

Note that even without casting the constness away, one might cause invalidation of a pointer by calling c_str, saving the pointer and then calling non-const method (which will cause write).
Another example without casting the constness away:

int main()
{
    string x = "hello";
    //copy constructor has been called here.
    string y(x);

    //y[0] = 'p';

    //c_str return const char*, but this usage is quite popular.
    const char* temp = y.c_str();

    y[0] = 'p';

    //Now we expect "pello" because the standart says the pointer points to the actual data
    //but we will get "hello"
    cout << "temp = " << temp << endl; 



    return 0;
}

g++ 是否满足 std::string C++11 要求

问题描述

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