函数签名指示返回类型为void *，但返回其他类型的指针

Question

I'm confused about the following code, with the function signature indicating the function returns a void pointer. The type actually returned however, is BlockInfo*. The function searches the first free memory block of size reqSize in a list. The whole program compiles and runs correctly.

static void * searchFreeList(size_t reqSize) {
  BlockInfo* freeBlock;

  freeBlock = FREE_LIST_HEAD;
  while (freeBlock != NULL){
    if (SIZE(freeBlock->sizeAndTags) >= reqSize) {
      return freeBlock;
    } else {
      freeBlock = freeBlock->next;
      }
  }
  return NULL;

}

My questions are: (1) Why is return freeBlock ,which returns type BlockInfo* a valid statement. (2) For a function that returns void pointer, when its return value is assigned to another pointer variable, eg:

int * ptr;  // Or double* ptr, etc.
ptr = searchFreeList(someSize);

Are these valid assignments?

Thanks a lot!

Answer 1

Why is return freeBlock,which returns type BlockInfo* a valid statement.

void * is a generic pointer type and is guaranteed to hold any object pointer type. There is an implicit conversion between all object pointer types to void * so no cast is even needed in the return statement.

int * ptr; // Or double* ptr, etc. ptr = searchFreeList(someSize); Are these valid assignments?

Yes, they are valid assignments but dereferencing the pointer may be then undefined behavior if the underlying type was a different type.

For example:

char a = 42;
void *p = &a;
int *q = p;

*q;  // undefined behavior

Answer 2

In C, void * is a generic pointer type. It can point to any pointer type. So,

ptr = searchFreeList(someSize);  

is a valid assignment.

Answer 3

A void pointer can store address of any type and in this case it is storing address of type BlockInfo . So when a function returns a pointer of type BlockInfo it is totally fine because the prototype doesn't care for the type in this case.

The second part of your question where you ask whether

int *ptr = searchFreeList(someSize);

is valid. Yes

int *p = malloc(sizeof(int) *2);

is valid , here malloc() returns void *

The result will be that pointer of type int pointing to some struct and dereferencing it will be UB