


C++ compilation error: Cannot call member function converted from volatile type, how to deal with it?
C is a strongly typed language that strictly limits the type conversion of variables. However, in some cases, we may need to perform type conversion on volatile type objects. Especially in embedded development, we often need to access hardware. Registers, and these registers are usually of volatile type. However, because volatile type objects have special semantics, the C compiler will impose some special restrictions on them, which leads to the error "Cannot call member functions converted from volatile types". This article will explain the cause of this error and how to deal with it.
First, let’s look at the semantics of volatile types. In C, the role of the volatile keyword is to tell the compiler that the value of this variable may be modified outside the program, so the compiler cannot optimize it and must ensure that its value is re-read every time it is accessed. Specifically, volatile objects have the following characteristics:
- The value of a volatile object can be modified outside the program, such as hardware interrupts, multi-threading, etc.
- Every time a volatile object is accessed, its value must be re-read, and the cached value in the register cannot be used directly.
- Access to volatile objects cannot be reordered or optimized and must be performed in the order in the program.
Under this semantics, we can use volatile type objects to represent hardware registers. It should be noted that volatile type objects cannot be converted to and from non-volatile type objects, because this will destroy its special semantics. For example, the following code is wrong:
int x = 0; volatile int &y = x; // 复制x的地址,但y是volatile类型 x = 1; // OK,修改x的值 y = 2; // OK,修改x的值,但要重新读取其值 int z = y; // 错误,不能读取volatile对象的值 int &u = y; // 错误,不能将volatile类型的引用转换为非volatile类型
In the above code, we try to convert the non-volatile type variable x into a volatile type reference y, which is wrong. Although by doing this we can modify the value of x via y and re-read its value with each modification, we cannot read the value of y like a normal integer because this would violate the semantics of the volatile type.
Further, let us consider a more complex situation, that is, calling a member function on an object of volatile type. For example, we can declare a member function of an object as a volatile type, so that the visibility of its member variables can be guaranteed when it is called. However, the C compiler does not allow conversion from volatile types to non-volatile types, so the compilation error "Cannot call member function converted from volatile type" will occur. For example:
class MyClass { public: volatile int x; volatile void func() { x = x + 1; } }; int main() { MyClass obj; obj.func(); // 错误,不能从volatile类型转换为非volatile类型 return 0; }
In the above code, we define a MyClass class, where x is an integer of volatile type, and func() is a member function of volatile type, which means to perform an auto-increment operation on x . In the main() function, we create a MyClass object obj and try to call its member function func(). However, this will cause the compilation error "Cannot call a member function converted from volatile type". This is because, in C, member functions are treated as ordinary functions with a hidden this pointer parameter, so when calling the member function, converting the this pointer from a non-volatile type to a volatile type is not allowed.
So, how should we deal with this compilation error? There are two ways to solve this problem. The first method is to declare the parameters of the member function as volatile types so that the compiler will not report an error. For example:
class MyClass { public: volatile int x; void func(volatile MyClass *thisptr) { thisptr->x = thisptr->x + 1; } }; int main() { MyClass obj; obj.func(&obj); // OK,将this指针转换为volatile类型 return 0; }
In the above code, we declare the parameter thisptr of the func() function as a MyClass pointer of volatile type, so that when calling it, the this pointer can be converted from a non-volatile type to a volatile type. . While this approach can solve the problem, it makes the code verbose and is therefore not very commonly used.
The second method is to use type erasure technology to convert the this pointer of the member function into a void pointer, so that the compiler's restrictions on volatile types can be bypassed. For example:
class MyClass { public: volatile int x; void func() { volatile void *vthis = static_cast<volatile void *>(this); volatile MyClass *vptr = static_cast<volatile MyClass *>(vthis); vptr->x = vptr->x + 1; } }; int main() { MyClass obj; obj.func(); // OK,使用类型擦除将this指针转换为volatile类型 return 0; }
In the above code, we use static_cast to convert this pointer to a void pointer first, and then convert it to a volatile MyClass pointer, so that we can obtain a volatile this pointer. Although this approach can solve the problem, it requires understanding how to use type erasure techniques and may affect the readability and maintainability of the code.
To sum up, the C compilation error "Cannot call a member function converted from a volatile type" is caused by the compiler's special restrictions on volatile types. In order to solve this compilation error, we can declare the parameters of the member function as volatile types, or use type erasure technology to convert the this pointer of the member function into a void pointer. No matter which method is used, you need to pay attention to the semantics of volatile types to prevent converting volatile objects into non-volatile objects and vice versa, which may lead to incorrect results.
The above is the detailed content of C++ compilation error: Cannot call member function converted from volatile type, how to deal with it?. For more information, please follow other related articles on the PHP Chinese website!

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