为什么C中需要挥发性?它的用途是什么?它会做什么?


当前回答

在Dennis Ritchie设计的语言中,除了地址未被获取的自动对象外,对任何对象的每次访问都表现为计算对象的地址,然后在该地址上读写存储。这使得该语言非常强大,但严重限制了优化机会。

While it might have been possible to add a qualifier that would invite a compiler to assume that a particular object wouldn't be changed in weird ways, such an assumption would be appropriate for the vast majority of objects in C programs, and it would have been impractical to add a qualifier to all the objects for which such assumption would be appropriate. On the other hand, some programs need to use some objects for which such an assumption would not hold. To resolve this issue, the Standard says that compilers may assume that objects which are not declared volatile will not have their value observed or changed in ways that are outside the compiler's control, or would be outside a reasonable compiler's understanding.

Because various platforms may have different ways in which objects could be observed or modified outside a compiler's control, it is appropriate that quality compilers for those platforms should differ in their exact handling of volatile semantics. Unfortunately, because the Standard failed to suggest that quality compilers intended for low-level programming on a platform should handle volatile in a way that will recognize any and all relevant effects of a particular read/write operation on that platform, many compilers fall short of doing so in ways that make it harder to process things like background I/O in a way which is efficient but can't be broken by compiler "optimizations".

其他回答

在Dennis Ritchie设计的语言中,除了地址未被获取的自动对象外,对任何对象的每次访问都表现为计算对象的地址,然后在该地址上读写存储。这使得该语言非常强大,但严重限制了优化机会。

While it might have been possible to add a qualifier that would invite a compiler to assume that a particular object wouldn't be changed in weird ways, such an assumption would be appropriate for the vast majority of objects in C programs, and it would have been impractical to add a qualifier to all the objects for which such assumption would be appropriate. On the other hand, some programs need to use some objects for which such an assumption would not hold. To resolve this issue, the Standard says that compilers may assume that objects which are not declared volatile will not have their value observed or changed in ways that are outside the compiler's control, or would be outside a reasonable compiler's understanding.

Because various platforms may have different ways in which objects could be observed or modified outside a compiler's control, it is appropriate that quality compilers for those platforms should differ in their exact handling of volatile semantics. Unfortunately, because the Standard failed to suggest that quality compilers intended for low-level programming on a platform should handle volatile in a way that will recognize any and all relevant effects of a particular read/write operation on that platform, many compilers fall short of doing so in ways that make it harder to process things like background I/O in a way which is efficient but can't be broken by compiler "optimizations".

Volatile告诉编译器你的变量可以通过其他方式被改变,而不是通过访问它的代码。例如,它可能是一个I/ o映射的内存位置。如果在这种情况下没有指定这一点,一些变量访问可以被优化,例如,它的内容可以保存在寄存器中,并且内存位置不会再次读入。

维基上有关于volatile的所有信息:

Volatile(计算机编程)

Linux内核的文档也对volatile做了一个很好的标记:

为什么不应该使用“volatile”类型类

volatile的边缘用法如下。假设你想计算一个函数f的数值导数:

double der_f(double x)
{
    static const double h = 1e-3;
    return (f(x + h) - f(x)) / h;
}

问题是由于舍入误差,x+h-x通常不等于h。想想看:当你减去非常接近的数字时,你会丢失很多有效的数字,这可能会破坏导数的计算(想想1.00001 - 1)

double der_f2(double x)
{
    static const double h = 1e-3;
    double hh = x + h - x;
    return (f(x + hh) - f(x)) / hh;
}

但是根据您的平台和编译器开关的不同,该函数的第二行可能会被积极优化的编译器删除。所以你可以写

    volatile double hh = x + h;
    hh -= x;

强制编译器读取包含hh的内存位置,从而丧失最终的优化机会。

我会提到另一个挥发物很重要的场景。

假设您为更快的I/O对文件进行内存映射,并且该文件可以在幕后更改(例如,该文件不在您的本地硬盘驱动器上,而是通过网络由另一台计算机提供)。

如果您通过指向非易失性对象的指针(在源代码级别)访问内存映射文件的数据,那么编译器生成的代码可以多次获取相同的数据,而您却不知道它。

如果该数据碰巧发生了变化,您的程序可能会使用两个或多个不同版本的数据,并进入不一致的状态。如果程序处理不受信任的文件或来自不受信任位置的文件,这不仅会导致程序的逻辑不正确行为,而且还会导致可利用的安全漏洞。

如果您关心安全性,这是一个需要考虑的重要场景。