C++ 从“void*”到“unsigned char*”的无效转换

Invalid conversion from ‘void*’ to ‘unsigned char*’

提问by jwbensley

I have the following code;

我有以下代码；

void* buffer = operator new(100);
unsigned char* etherhead = buffer;

I'm getting the following error for that line when trying to compile;

尝试编译时，该行出现以下错误；

error: invalid conversion from ‘void*' to ‘unsigned char*'

Why do I get that error, I thought a void was "type-less" so it can point at anything, or anything can point to it?

为什么我会收到这个错误，我认为 void 是“无类型的”，所以它可以指向任何东西，或者任何东西都可以指向它？

回答by josephthomas

You need to cast as you can not convert a void* to anything without casting it first.

您需要进行强制转换，因为如果不先进行转换，就无法将 void* 转换为任何内容。

You would need to do

你需要做

unsigned char* etherhead = (unsigned char*)buffer;

(although you could use a static_castalso)

（虽然你也可以使用 a static_cast）

To learn more about void pointers, take a look at 6.13 — Void pointers.

要了解有关空指针的更多信息，请查看6.13 — 空指针。

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The "type-less" state of void*only exist in C, not C++ with stronger type-safety.

“无类型”状态void*只存在于 C 中，而不存在于具有更强类型安全性的 C++ 中。

回答by CB Bailey

A void*might point at anything and you can convert a pointer to anything else to a void*without a cast but you have to use a static_castto do the reverse.

Avoid*可能指向任何东西，您可以将指向任何其他东西的指针转换为 avoid*而无需强制转换，但您必须使用 astatic_cast来做相反的事情。

unsigned char* etherhead = static_cast<unsigned char*>(buffer);

If you want a dynamically allocated buffer of 100 unsigned charyou are better off doing this and avoiding the cast.

如果您想要动态分配的 100 缓冲区，unsigned char最好这样做并避免强制转换。

unsigned char* p = new unsigned char[100];

回答by mjfgates

You can convert any pointer to a void *, but you can't convert void * to anything else without a cast. It might help to imagine that "void" is the base class for EVERYTHING, and "int" and "char" and whatnot are all subclasses of "void."

您可以将任何指针转换为 void *，但不能在没有强制转换的情况下将 void * 转换为任何其他指针。想象一下“void”是所有东西的基类可能会有所帮助，而“int”和“char”等等都是“void”的子类。

回答by Philipp

Here's a bit of lateral thinking: Whenever you think you need casts or pointers, think again. If all you need is 100 unsigned bytes of memory, use

这里有一点横向思考：每当你认为你需要强制转换或指针时，再想一想。如果您只需要 100 个无符号字节的内存，请使用

std::array<unsigned char, 100> data;

or

或者

unsigned char data[100];

If the size is not constant, use a vector:

如果大小不恒定，请使用向量：

std::vector<unsigned char> data(size);

Raw pointers, the newoperator, and casts are unsafe, hard to get right and make your program harder to understand. Avoid them if possible.

原始指针、new操作符和强制转换是不安全的，难以正确理解并且使您的程序更难理解。如果可能，请避免使用它们。

回答by Spike

C++ is designed to be more type safe than C. If this is C code, it may be OK, but also depends on what compiler you are using right now.

C++ 被设计为比 C 更安全。如果这是 C 代码，它可能没问题，但也取决于您现在使用的编译器。

Also, technically, "extern "C" int *" and "int *" are different types... (like solaris compiler will pick this out)

此外，从技术上讲，"extern "C" int *" 和 "int *" 是不同的类型......（就像solaris编译器会挑选出来）

I would suggest you using C++ style cast instead of C cast. There are more descriptions here:

我建议您使用 C++ 样式转换而不是 C 转换。这里有更多描述：

Regular cast vs. static_cast vs. dynamic_cast.

常规转换 vs. static_cast vs. dynamic_cast。

回答by Erwald

void *pt; 

pt=(void*)&i; 
pt=(void*)&dbl; 

Here's how I would do it.

这是我将如何做到的。