VirtualAlloc失败

时间:2015-04-26 04:46:00

标签: c++ windows memory virtualalloc

我正在尝试使用VirtualAlloc来保留和提交一块内存,然后再次扩展该块。不幸的是,尽管VirtualQuery声称所请求的地址范围是空闲的,但它返回NULL且错误为ERROR_INVALID_ADDRESS。这是我的代码:

void* allocation = VirtualAlloc(NULL, 4096, MEM_RESERVE | MEM_COMMIT, PAGE_READWRITE);
void* desiredNextAllocation = (char*)allocation + 4096;
MEMORY_BASIC_INFORMATION info;
size_t memory_info = VirtualQuery(desiredNextAllocation, &info, sizeof(info));
void* extended = VirtualAlloc(desiredNextAllocation, 4096, MEM_RESERVE | MEM_COMMIT, PAGE_READWRITE);

第一个分配返回0x00000000000d0000。对VirtualQuery的调用会在'info'中生成以下数据:

    BaseAddress 0x00000000000d1000  void *
    AllocationBase  0x0000000000000000  void *
    AllocationProtect   0x00000000  unsigned long
    RegionSize  0x00000000000ff000  unsigned __int64
    State   0x00010000  unsigned long
    Protect 0x00000001  unsigned long
    Type    0x00000000  unsigned long

我认为这意味着有0xff可用页面从0xd1000开始处于MEM_FREE状态。那么为什么我尝试在0xd1000提交页面失败?

我正在运行Windows 7,这是一个64位版本。

我已经阅读了几篇关于VirtualAlloc的StackOverflow帖子,但他们似乎都暗示这段代码应该像我对文档的理解一样工作。

2 个答案:

答案 0 :(得分:4)

如果要为分配指定连续页面,则需要将分配地址空间与分配内存分开以备份它。记住这一点,我们可以实现类似这样的代码:

epw-title

示例结果:

#include <windows.h>
#include <iostream>
#include <iomanip>

std::ostream &operator<<(std::ostream &os, MEMORY_BASIC_INFORMATION const &mi) {
    return os   << std::setw(20) << "Allocation Base: " << mi.AllocationBase << "\n"
                << std::setw(20) << "BaseAddress: " << mi.BaseAddress << "\n"
                << std::setw(20) << "Protection: " << mi.Protect << "\n"
                << std::setw(20) << "Region size: " << mi.RegionSize;
}

void show_page(void *page) {
    MEMORY_BASIC_INFORMATION info;

    VirtualQuery(page, &info, sizeof(info));
    std::cout << info << "\n\n";
}

static const int page_size = 4096;

void *alloc_page(char *address) {

    void *ret = VirtualAlloc(address, page_size, MEM_COMMIT, PAGE_READWRITE);
    show_page(ret);
    return ret;
}

int main() {
    static const int region_size = 65536;

    char * alloc = static_cast<char *>(VirtualAlloc(NULL, region_size, MEM_RESERVE, PAGE_READWRITE));

    for (int i = 0; i < 4; i++)
        alloc_page(alloc + page_size * i);
}

如您所见,所有分配现在都成功了。另外:当您保留地址空间时,您可以分配的最小大小为64K(如上所示)。您应该通过调用Allocation Base: 00000000000C0000 BaseAddress: 00000000000C0000 Protection: 4 Region size: 4096 Allocation Base: 00000000000C0000 BaseAddress: 00000000000C1000 Protection: 4 Region size: 4096 Allocation Base: 00000000000C0000 BaseAddress: 00000000000C2000 Protection: 4 Region size: 4096 Allocation Base: 00000000000C0000 BaseAddress: 00000000000C3000 Protection: 4 Region size: 4096 ,并使用GetSystemInfo结构中的dwPageSizedwAllocationGranularity来获取页面大小和最小区域大小。

答案 1 :(得分:1)

来自documentation for VirtualAlloc

  

如果正在保留内存,则将指定的地址向下舍入到分配粒度的最接近的倍数。

在这种情况下,地址0xd1000向下舍入到地址0xd0000,该地址已被保留,因此无效。