process-請大家幫我看看兩段代碼的區別真想不明白了

請大家幫我看看兩段代碼的區別真想不明白了

有兩段代碼 代碼一是單獨寫成函數來調用的， 我把他們合成一個主函數的代碼 就出錯  好像是變量遞增的問題，我怎麼也想不通，麻煩大家了。

代碼一  別人寫的：
#include
#include
#include
#include
#pragma comment(lib,"Psapi.lib")

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

    // 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];
            if (GetModuleBaseName(hProcess, hMods[i], szModName,sizeof(szModName) / sizeof(TCHAR)))
            {
if(strcmp(szModName,"sfc.dll")==0) 
{
printf("%s PID:%d\n",szModName,processID);
break;
}
            }
        }
    }
    CloseHandle( hProcess );
    return 0;
}

int main( void )
{
    DWORD aProcesses[1024]; 
    DWORD cbNeeded; 
    unsigned int i;

    // Get the list of process identifiers.
    if ( !EnumProcesses( aProcesses, sizeof(aProcesses), &cbNeeded ) ) return 1;

    // Print the names of the modules for each process.
    for ( i = 0; i < cbNeeded / sizeof(DWORD); i++ )
    {
        FindModules( aProcesses[i] );
    }

    return 0;
}

代碼2：我想合到一個函數中：

#include
#include
#include
#pragma comment(lib,"Psapi.lib")

int main( void )
{
    DWORD aProcesses[1024]; 
    DWORD cbNeeded; 
    unsigned int i;

HMODULE hMods[1024];
HANDLE hProcess;
DWORD cbNeeded2;
unsigned int ii;

    // Get the list of process identifiers.
    if ( !EnumProcesses( aProcesses, sizeof(aProcesses), &cbNeeded ) ) return 1;

    // Print the names of the modules for each process.
    for ( i = 0; i < cbNeeded / sizeof(DWORD); i++ )
    {
        // 錯誤下面這句，分開些函數就沒問題 i都等於0，寫在一個函數裡 就不行了，就返回2，必須把i加1，跳過過pid為0才行.
hProcess = OpenProcess( PROCESS_QUERY_INFORMATION | PROCESS_VM_READ, FALSE, aProcesses[i+1] );//我直接用aProcesses[i]怎麼不行啊？
if (NULL == hProcess) return 2;

   // Get a list of all the modules in this process.
    if( EnumProcessModules(hProcess, hMods, sizeof(hMods), &cbNeeded2))
    {
        for ( ii = 0; ii < (cbNeeded2 / sizeof(HMODULE)); ii++ )
        {
            TCHAR szModName[MAX_PATH];
            if (GetModuleBaseName(hProcess, hMods[ii], szModName,sizeof(szModName) / sizeof(TCHAR)))
            {
                if(strcmp(szModName,"sfc.dll")==0) 
                {
                    printf("%s  PID:%d\n",szModName,aProcesses[i+1]);
                    break;
                }
            }
        }
        CloseHandle( hProcess );
    }
}

    return 0;
}

最佳回答：

基礎太差了 單獨寫成函數是循環來調用，遇到pid=0的雖然失敗 但循環下一個， 合寫在一個主函數的時候沒仔細看代碼 直接把上面的return 給挪下來了，這就出現了id寫成aProcesses[i] 會失敗返回 寫成aProcesses[i+1]就跳過pid=0了。或者把return 換成continue就ok了