在VS 2010中,static_cast无法从void *转换为C ++中的size_t错误

时间:2012-04-30 09:46:09

标签: c++

我有以下代码,我收到错误

  

错误C2036:'void *':未知大小
  错误C2440:'static_cast':无法从'void *'转换为'size_t'

在第

void *addr = static_cast <void*> (static_cast <size_t> (mem + bytesAlreadyAllocated));

我的问题是为什么我会遇到错误以及如何摆脱这些错误?

感谢您的帮助。

class MemoryChunk {
public:
    MemoryChunk (MemoryChunk *nextChunk, size_t chunkSize);
    ~MemoryChunk() {delete mem; }

    inline void *alloc (size_t size);
    inline void free (void* someElement);

    // Pointer to next memory chunk on the list.
    MemoryChunk *nextMemChunk() {return next;}
    // How much space do we have left on this memory chunk?
    size_t spaceAvailable() { return chunkSize - bytesAlreadyAllocated; }

    // this is the default size of a single memory chunk.
    enum { DEFAULT_CHUNK_SIZE = 4096 };
private:

    // The MemoryChunk class is a cleaner version of NextOnFreeList. It separates the next pointer from
    // the actual memory used for the allocated object. It uses explicit next and mem pointers, with no 
    // need for casting.

    MemoryChunk *next;
    void *mem;

    // The size of a single memory chunk.
    size_t chunkSize;
    // This many bytes already allocated on the current memory chunk.
    size_t bytesAlreadyAllocated;
};

MemoryChunk::MemoryChunk(MemoryChunk *nextChunk, size_t reqSize) {
    chunkSize = (reqSize > DEFAULT_CHUNK_SIZE) ? reqSize : DEFAULT_CHUNK_SIZE;
    next = nextChunk;
    bytesAlreadyAllocated = 0;
    mem = new char [chunkSize];
}

void* MemoryChunk :: alloc (size_t requestSize) {
    void *addr = static_cast <void*> (static_cast <size_t> (mem + bytesAlreadyAllocated));
    bytesAlreadyAllocated += requestSize;
    return addr;
}

inline void MemoryChunk :: free (void *doomed) {}

1 个答案:

答案 0 :(得分:4)

表达式:

static_cast <size_t> (mem + bytesAlreadyAllocated)

使用未定义大小(void)的类型应用偏移量。由于void没有大小,程序形成不良。

char*是适合您在此方案中使用的指针。例如:

`char* mem;`
 // and
 char* addr(mem + bytesAlreadyAllocated);

<强>更新

所以在以下程序中:

#include <iostream>

int main(int argc, const char* argv[]) {
    const int array[3] = {-1, 0, 1};
    std::cout << *(array + 0) << ", "
      << *(array + 1) << ", " << *(array + 2) << "\n";
    return 0;
}

输出为-1, 0, 1。根据数组元素的大小和类型应用元素偏移。使用void - 尺寸不合适。