c - 失去不稳定资格

Question

我有一个代码在一些安全关键的汽车模块中运行。以下是代码的粗略估计:

下面的代码是模块“主模块”的一部分，它拥有 volatile 变量/数组“x_ast”

Main Module.c

//The structure x contains the major data that needs to be stored in case of a crash event. i.e. a real car crash
// x contains data such a vehicle speed, environment data, sensor data, CAN related data,etc. Basically x itself has lot of structures inside it.


typedef struct x x_tst;
volatile x_tst x_ast[5];
//x_ast is being used in realtime and background functions, considering proper interrupt disabling and enabling.

下面的代码是模块“DependentModule”的一部分，它支持共享缓冲区“x_ast”。

DependentModule.c

extern volatile x_tst x_ast[5];


//x_ast is owned by a separate module
//Now I need to share this buffer "x_ast" to a different module that will inturn fill data in it for some conditions. Say for a particular condition that is activated, the buffer "x_ast" will be shared across to a module "Conditional Data Record". 
//The base address of first indexed buffer "x_ast[1]" is provided to "Conditional Data Record" for access. 
//The main module will not access the buffer that is shared, once the "Conditional Data Record" is activated/requested. 

// Below is a mechanism to share the buffer "x_ast" to module "Conditional Data Record".

// This API will be called by module - "Conditional Data Record" once it is activated. It is ensured that this takes place only during intialization of whole system, and no other time.

boolean GetConditionalBuffer(uint8 **PtrToBuffer)
{
  boolean BufferShared = FALSE;
  void* Temp_ptr;
  *PtrToBuffer = NULL;

// if module  "Conditional Data Record" is activated? then share the buffer "x_ast", else do not share the buffer.

if(Conditional Data Record == activated) {
   Temp_ptr = (x_tst*)&x_ast[1];
   *PtrToBuffer = Temp_ptr;
   BufferShared = TRUE;
}

return BufferShared;

}



Referring to the line of code:
Temp_ptr = (x_tst*)&x_ast[1];

上面的代码行(Tempptr = (x_tst*)&x_ast[1];)抛出警告“Msg(7:0312)危险的指针转换导致失去 volatile 资格.”上述警告是强制警告，因此有必要解决它。

我确实明白，我将 volatile 变量的地址分配给 void 指针，这会导致失去 volatile 资格。我尝试了不同的方法试图解决该警告，但无法得到结论性的方法。

有什么办法，我可以修改代码并删除此警告，或者可以绕过此警告。

Answer 1

如果您将值分配给 volatile 对象或读取 volatile 对象的值而不使用 volatile 限定指针，则会出现未定义的行为。

编译器必须以比非 volatile 对象更严格的方式对待 volatile 对象(这意味着将非 volatile 对象视为 volatile 对象很好，将 volatile 对象视为非 volatile 对象可能会产生不好的结果)结果)。

将 volatile 对象的地址转换为非 volatile 指针会让您面临严重的风险。不要这样做。无论谁使用该 void* 指针，都有调用未定义行为的巨大风险。例如，仅使用 memcpy 复制该 volatile 数组是未定义的行为。任何事情，通常是坏事，都可能发生。

将该函数声明为

boolean GetConditionalBuffer(volatile x_tst **PtrToBuffer)

因为 volatile x_tst* 是它存储到 PtrToBuffer 中的内容。为什么要丢弃类型信息？我实际上会将其更改为

volatile x_tst* GetConditionalBuffer (void);

这使得可怜的开发人员的大脑变得更容易，并且使该功能更易于使用。