C++：malloc：错误：从“void*”到“uint8_t*”的无效转换
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C++: malloc : error: invalid conversion from ‘void*’ to ‘uint8_t*’
提问by olidev
I got this problem:
我遇到了这个问题：
invalid conversion from ‘void*' to ‘uint8_t*'
从“void*”到“uint8_t*”的无效转换
When doing this:
这样做时：
int numBytes;
uint8_t *buffer;
buffer=malloc(numBytes); //error here, why?
or must I have to put it like this?
还是我必须这样说？
buffer=malloc(numBytes);
Please explain this.
请解释一下。
回答by Oliver Charlesworth
You cannot implicitly cast from void *in C++ (unlike C in this respect). You could do:
您不能void *在 C++ 中隐式转换（在这方面与 C 不同）。你可以这样做：
buffer = static_cast<uint8_t *>(malloc(numBytes));
but really, you should just be using new/deleteinstead of malloc/free!
但实际上，您应该只使用new/delete而不是malloc/ free！
回答by Sam Dufel
Malloc returns a void pointer; when you use it, you have to cast the return value to a pointer to whatever datatype you're storing in it.
Malloc 返回一个空指针；当您使用它时，您必须将返回值转换为指向您存储在其中的任何数据类型的指针。
buffer = (uint8_t *) malloc(numBytes);
回答by Nils Pipenbrinck
In C++ it is not allowed to simply assign a pointer of one type to a pointer of another type (as always there are exception to the rule. It is for example valid to assign a any pointer to a void pointer.)
在 C++ 中，不允许简单地将一种类型的指针分配给另一种类型的指针（因为规则总是有例外。例如，将 any 指针分配给 void 指针是有效的。）
What you should do is to cast your void pointer to a uint8_t pointer:
您应该做的是将 void 指针转换为 uint8_t 指针：
buffer = (uint8_t *) malloc (numBytes);
Note: This is only necessary in C++, in C it was allowed to mix and match pointers. Most C compilers give a warning, but it is valid code.
注意：这仅在 C++ 中是必需的，在 C 中允许混合和匹配指针。大多数 C 编译器都会发出警告，但它是有效的代码。
Since you're using C++ you could also use new and delete like this:
由于您使用的是 C++，您还可以像这样使用 new 和 delete：
buffer = new uint8_t[numBytes];
and get rid of your buffer using:
并使用以下方法摆脱缓冲区：
delete[] buffer;
In general you shouldn't use malloc and free unless you have to interface with c libraries.
一般来说，除非必须与 c 库接口，否则不应使用 malloc 和 free。
