将变量强制转换为 (void *) 有什么效果？
在 C 程序中，我看到以下语句：
memcpypgm2ram((void*)&AppConfig.MyMACAddr, (ROM void*)SerializedMACAddress, sizeof(AppConfig.MyMACAddr));
(void*) case 的作用是什么？这是为 Microchip C30 编译器编写的。
AppConfig 的定义如下：
APP_CONFIG AppConfig; // APP_CONFIG is obviously a structure...
SerializedMACAddress 的定义如下：
static ROM BYTE SerializedMACAddress[6] = {MY_DEFAULT_MAC_BYTE1, MY_DEFAULT_MAC_BYTE2, MY_DEFAULT_MAC_BYTE3, MY_DEFAULT_MAC_BYTE4, MY_DEFAULT_MAC_BYTE5, MY_DEFAULT_MAC_BYTE6};
编辑：我应该在之前声明这一点，但 memcpypgm2ram 定义为： #define memcpypgm2ram(a,b,c) memcpy(a,b,c)
所以基本上，void *memcpy(void *dest, const void *src, size_t n);
In a C program I see the following statement:
memcpypgm2ram((void*)&AppConfig.MyMACAddr, (ROM void*)SerializedMACAddress, sizeof(AppConfig.MyMACAddr));
What does the (void*) case do? This is written for the Microchip C30 compiler.
AppConfig is defined like this:
APP_CONFIG AppConfig; // APP_CONFIG is obviously a structure...
SerializedMACAddress is defined like this:
static ROM BYTE SerializedMACAddress[6] = {MY_DEFAULT_MAC_BYTE1, MY_DEFAULT_MAC_BYTE2, MY_DEFAULT_MAC_BYTE3, MY_DEFAULT_MAC_BYTE4, MY_DEFAULT_MAC_BYTE5, MY_DEFAULT_MAC_BYTE6};
EDIT: I should have stated this before but memcpypgm2ram is defined as:
#define memcpypgm2ram(a,b,c) memcpy(a,b,c)
so basically, void *memcpy(void *dest, const void *src, size_t n);
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void* 是通用数据指针类型，当用作参数类型时，表示函数在“裸”内存块上工作。它不能被取消引用。
任何其他数据指针类型都可以隐式转换为
void*，因此显式转换可能是错误的（不必要），或者是损坏编译器的解决方法，或者是转换为unsigned 的简写char *（在这种情况下，这是接口损坏的解决方法）。void*is the universal data pointer type, that, when used as an argument type, denotes that a function works on "bare" memory blocks. It cannot be dereferenced.Any other data pointer type can be implicitly converted to
void*, so the explicit cast is probably either wrong (unnecessary), or a workaround for a broken compiler, or a shorthand to cast tounsigned char *(in which case it's a workaround for a broken interface).强制转换
void *将某种类型的指针转换为通用指针。The cast
void *converts a pointer of some type into a Generic Pointer.memcpypgm2ram 函数不需要特定类型。它应该用以下原型定义： memcpypgm2ram(void* p1, void* p2, int n);
“void *”是对通用指针类型的强制转换。
该函数仅接受任意类型的两个指针，并将 n 个字节（在您的情况下为 n=sizeof(AppConfig.MyMACAddr)）从一个地址复制到另一个地址。
The memcpypgm2ram function does not need a specific type. It should be defined with the following prototype : memcpypgm2ram(void* p1, void* p2, int n);
"void *" is a cast to a generic pointer type.
The function just takes two pointers of any type and copy n bytes (in your case n=sizeof(AppConfig.MyMACAddr)) from one address to the other.
memcpypgm2ram 可能采用 void 指针作为参数。 C 的类型足够强，可以识别 &AppConfig.MyMACAddr 的类型是 (MACAddr*)，并且如果您不将其转换为 void*，则会发出编译时错误。
关键是 memcpypgm2ram 是一个可以处理内存中保存的任何字节的函数，因此它不接受强类型指针作为参数。
memcpypgm2ram presumably takes a void pointer as it's argument. C is strongly enough typed to recognize that the type of &AppConfig.MyMACAddr is (MACAddr*) and will emit a compile-time error if you don't cast it as a void*.
The point is that memcpypgm2ram is a function that works on any bytes held in memory, so it won't accept strongly typed pointers as arguments.
在这种情况下，可以使用强制转换为 (void *) 来消除编译器的警告。
in this case, casting to (void *) may be used to get rid of compiler's warning.
在没有看到更多上下文的情况下，无法说明在这种特定情况下使用
(void *)转换的意义是什么（memcpypgm2ram是如何声明的？）。在 C 语言（以及 C++）中，指针类型可以隐式转换为
void *类型，这意味着在指针转换中通常没有理由使用显式强制转换为void * 类型。在您的示例中，所有转换看起来都是指针转换，因此，考虑到我所说的，不需要显式转换为void *。另一种可能性是原始指针类型是 const 限定的，因此使用强制转换为
void *来删除 const 限定。但是，我在您提供的内容中没有看到任何 const 限定，这再次意味着很可能不需要强制转换为void *。我的猜测是，无论是谁把它放在那里，都是“以防万一”，没有任何真正的原因。简而言之，
(void *)转换将指针转换为void *类型。但由于该转换无论如何都会隐式发生，因此完全没有必要进行强制转换（假设使用 void * 参数声明 memcpypgm2ram）。There's no way to say what was the point of using
(void *)cast in this specific case without seeing more context (how ismemcpypgm2ramdeclared?).In C language (as well as in C++) pointer types are implicitly convertible to
void *type, which means that in pointer conversions there's usually no reason to use an explicit cast tovoid *type. It looks like in your example all conversions are pointer conversions, so, taking into account what I said, the explicit cast tovoid *is not required.Another possibility though is that the original pointer types are const-qualified, so the cast to
void *was used to remove the const-qualification. However, I don't see any const-qualifications in what you provided, which means, again, that most likely that cast tovoid *is unnecessary. My guess would be that whoever put it there did it "just in case" for no real reason.In short, the
(void *)cast converts the pointers tovoid *type. But since that conversion would happen implicitly anyway, the cast is totally unnecessary (assumingmemcpypgm2ramis declared withvoid *parameters).