Conversion from 'void*' to pointer to non-'void' requires an explicit cast

今天在混编C和c++时遇到了这样的错误，是个小错误哦，贴出来不怕大家笑，希望可以对我这样的初学者有所帮助，解决方法，就是因为在C语言中可以把空指针付给一个非空直接而不用去详细指明类型，而C++中是强类型语言所以要声明合适的转换类型

ANSI C allowed conversion from a void pointer to a non-void-pointer without explicitly casting

 

In C you can convert void* to non-void* without witing the conversion operator, but in C++ you can't. For example, this code is OK in C:

Code:

int *arr=malloc(16);

but is'll cause the same error as you got in C++ (malloc returns void*, and is forced to be assigned to int*). In C++ you should write (not as C++'sish at it seems ):

Code:

int *arr=(int*)malloc(16);

just defined the type that correspond to your function's declaration using typedef.