空指针：C和C ++的区别 [英] void pointers: difference between C and C++

问题描述

我想了解的问候空指针C和C ++之间的差异。在C编译如下而不是C ++（所有编译与海湾合作委员会所做/ G ++ -ansi -pedantic -Wall）：

I'm trying to understand the differences between C and C++ with regards to void pointers. the following compiles in C but not C++ (all compilations done with gcc/g++ -ansi -pedantic -Wall):

int* p = malloc(sizeof(int));

由于的malloc 收益无效* ，其中C ++不允许分配到为int * 而C呢。

Because malloc returns void*, which C++ doesn't allow to assign to int* while C does.

不过，在这里：

void foo(void* vptr)
{
}


int main()
{
    int* p = (int*) malloc(sizeof(int));
    foo(p);
    return 0;
}

这两个C ++和C没有抱怨编译。为什么呢？

Both C++ and C compile it with no complain. Why?

K＆放大器; R2说：

K&R2 say:

不限指针的对象可以是
  转换为类型为void *无损失
  的信息。如果结果是
  转换回原始指针
  类型，原始指针是
  恢复。

Any pointer to an object may be converted to type void * without loss of information. If the result is converted back to the original pointer type, the original pointer is recovered.

这pretty资金全部有关于无效* 在C什么的C ++标准听写？

And this pretty sums all there is about void* conversions in C. What does C++ standard dictate?

推荐答案

在C，指针转换和从无效* 总是隐含的。

In C, pointer conversions to and from void* were always implicit.

在C ++中，从 T * 转换为无效* 是隐含的，但无效* 来什么都需要一个演员。

In C++, conversions from T* to void* are implicit, but void* to anything else requires a cast.

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