獲取Java進程加載的dll列表

Question

我一直花費大量的時間處理java進程錯誤文件，即hs_err_pid * .log。在這些文件中有一個標題爲Dynamic libraries的部分，它顯示了進程在失敗時當前加載的所有dll。

是否可以在正常運行條件下以編程方式訪問此信息？

Answer 1

在命令提示符下運行jdk \ bin路徑並在commond下運行。

JMAP（進程ID）

這將打印THT進程中加載​​的所有dll。 對編程不太確定。這工作。我用於自動化目的。

注：您的JDK和JRE版本應該是匹配（可壓性）

Answer 2

https://msdn.microsoft.com/en-us/library/windows/desktop/ms682621%28v=vs.85%29.aspx

#include <windows.h> 
#include <tchar.h> 
#include <stdio.h> 
#include <psapi.h> 

// To ensure correct resolution of symbols, add Psapi.lib to TARGETLIBS 
// and compile with -DPSAPI_VERSION=1 

int PrintModules(DWORD processID) 
{ 
    HMODULE hMods[1024]; 
    HANDLE hProcess; 
    DWORD cbNeeded; 
    unsigned int i; 

    // Print the process identifier. 

    printf("\nProcess ID: %u\n", processID); 

    // Get a handle to the process. 

    hProcess = OpenProcess(PROCESS_QUERY_INFORMATION | 
          PROCESS_VM_READ, 
          FALSE, processID); 
    if (NULL == hProcess) 
     return 1; 

    // Get a list of all the modules in this process. 

    if(EnumProcessModules(hProcess, hMods, sizeof(hMods), &cbNeeded)) 
    { 
     for (i = 0; i < (cbNeeded/sizeof(HMODULE)); i++) 
     { 
      TCHAR szModName[MAX_PATH]; 

      // Get the full path to the module's file. 

      if (GetModuleFileNameEx(hProcess, hMods[i], szModName, 
             sizeof(szModName)/sizeof(TCHAR))) 
      { 
       // Print the module name and handle value. 

       _tprintf(TEXT("\t%s (0x%08X)\n"), szModName, hMods[i]); 
      } 
     } 
    } 

    // Release the handle to the process. 

    CloseHandle(hProcess); 

    return 0; 
} 

int main(void) 
{ 

    DWORD aProcesses[1024]; 
    DWORD cbNeeded; 
    DWORD cProcesses; 
    unsigned int i; 

    // Get the list of process identifiers. 

    if (!EnumProcesses(aProcesses, sizeof(aProcesses), &cbNeeded)) 
     return 1; 

    // Calculate how many process identifiers were returned. 

    cProcesses = cbNeeded/sizeof(DWORD); 

    // Print the names of the modules for each process. 

    for (i = 0; i < cProcesses; i++) 
    { 
     PrintModules(aProcesses[i]); 
    } 

    return 0; 
}