我正在尝试创建一个程序，利用 MiniDumpWriteDump Windows API 转储另一个进程的内存。但是，它一直返回一个错误，提示 The handle is invalid。我对我的进程句柄非常有信心，因为我以前使用过 OpenProcess Windows API，所以我认为这就是我使用 CreateFileW 的方式。

我看过网上的例子，比如 this one但我什么也做不了。

到目前为止，这是我的代码:

package main

import (
    "fmt"
    "os"
    "strconv"
    "syscall"
    "unsafe"
)

var kernel32        = syscall.NewLazyDLL("kernel32.dll")
var procOpenProcess = kernel32.NewProc("OpenProcess")
var procCreateFileW = kernel32.NewProc("CreateFileW")
var procCloseHandle = kernel32.NewProc("CloseHandle")

var dbghelp               = syscall.NewLazyDLL("Dbghelp.dll")
var procMiniDumpWriteDump = dbghelp.NewProc("MiniDumpWriteDump")

func main() {
    fmt.Println("[ ] Starting Enum-DumpProcessMemory\n")

    pid, _ := strconv.Atoi(os.Args[1])
    fmt.Println("[-] PID            :", pid)

    processHandle, _, _ := procOpenProcess.Call(uintptr(0xFFFF), uintptr(1), uintptr(pid))
    fmt.Println("[-] Process Handle :", processHandle)

    path, _ := syscall.UTF16PtrFromString(os.Args[2])
    fileHandle, _, _ := procCreateFileW.Call(uintptr(unsafe.Pointer(path)), syscall.GENERIC_READ, syscall.FILE_SHARE_READ|syscall.FILE_SHARE_WRITE, 0, syscall.OPEN_EXISTING, syscall.FILE_ATTRIBUTE_NORMAL, 0)
    fmt.Println("[-] File Handle    :", fileHandle)

    ret, _, err := procMiniDumpWriteDump.Call(uintptr(processHandle), uintptr(pid), uintptr(fileHandle), 0x00061907, 0, 0, 0)

    if ret != 0 {
        fmt.Println("[+] Process memory dump successful")
    } else {
        fmt.Println("[x] Process memory dump not successful")
        fmt.Println(err)
    }
}

这是输出:

> .\Enum-DumpProcessMemory.exe 6892 C:\Users\user\Documents\dump.dmp
[ ] Starting Enum-DumpProcessMemory

[-] PID            : 6892
[-] Process Handle : 236
[-] File Handle    : 18446744073709551615
[x] Process memory dump not successful
The handle is invalid.

最佳答案

感谢 Jonathan Potter 的帮助，我的代码可以正常工作。问题是我试图为一个不存在的文件创建一个处理程序。

这是我的工作代码:

package main

import (
    "fmt"
    "os"
    "strconv"
    "syscall"
    "unsafe"
)

var kernel32        = syscall.NewLazyDLL("kernel32.dll")
var procOpenProcess = kernel32.NewProc("OpenProcess")
var procCreateFileW = kernel32.NewProc("CreateFileW")

var dbghelp               = syscall.NewLazyDLL("Dbghelp.dll")
var procMiniDumpWriteDump = dbghelp.NewProc("MiniDumpWriteDump")

func main() {
    fmt.Println("[ ] Starting Enum-DumpProcessMemory\n")

    pid, _ := strconv.Atoi(os.Args[1])
    fmt.Println("[-] PID            :", pid)

    processHandle, _, _ := procOpenProcess.Call(uintptr(0xFFFF), uintptr(1), uintptr(pid))
    fmt.Println("[-] Process Handle :", processHandle)

    if _, err := os.Stat(os.Args[2]); os.IsNotExist(err) {
        os.Create(os.Args[2])
    }
    path, _ := syscall.UTF16PtrFromString(os.Args[2])

    fileHandle, _, _ := procCreateFileW.Call(uintptr(unsafe.Pointer(path)), syscall.GENERIC_WRITE, syscall.FILE_SHARE_READ|syscall.FILE_SHARE_WRITE, 0, syscall.OPEN_EXISTING, syscall.FILE_ATTRIBUTE_NORMAL, 0)
    fmt.Println("[-] File Handle    :", fileHandle)

    ret, _, err := procMiniDumpWriteDump.Call(uintptr(processHandle), uintptr(pid), uintptr(fileHandle), 0x00061907, 0, 0, 0)

    if ret != 0 {
        fmt.Println("[+] Process memory dump successful")
    } else {
        fmt.Println("[x] Process memory dump not successful")
        fmt.Println(err)
    }
}

关于go - Windows API 错误 "The handle is invalid"，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/57086863/