void pointers: difference between C and C++

问题

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));

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

However, here:

void foo(void* vptr)
{
}

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

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

K&R2 say:

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.

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

回答1:

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

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

回答2:

C++ is more strongly-typed than C. Many conversions, specially those that imply a different interpretation of the value, require an explicit conversion. The new operator in C++ is a type-safe way to allocate memory on heap, without an explicit cast.

回答3:

It is useful to understand that pointer type conversions don't really require execution of extra CPU instructions. They are analysed during the compile time to understand the intensions of the developer. void * is an opaque pointer. All it says that the type of pointed object is unknown. C is weakly typed. It allows direct conversion between (void *) and any (T*) implicitly. C++ is strongly typed. A conversion from (void *) to (T*) wouldn't really make good case for a strongly typed language. But C++ had to stay backwards compatible with C, hence it had to allow such conversions. The guiding principle then is: explicit is better than implicit. Hence if you wish to convert an (void*) to some specific (T*) pointer, you need to explicitly write that in code. Conversion from (T*) to (void*) doesn't require explicit conversion since there is nothing much one can do on a (void*) pointer directly (one can call free() though). Hence (T*) to (void*) conversion is pretty much safe.

来源：https://stackoverflow.com/questions/1736833/void-pointers-difference-between-c-and-c

标签

c++

void-pointers