Confusion with ProcessBuilder in Java

Question

For the second time I had a problem with values extracted from system calls using ProcessBuilder.

@org.junit.Test
public void test() {
    Process process = null;
    ProcessBuilder pb = new ProcessBuilder("QQ.exe");
    pb.directory(new File("D:\\Program Files (x86)\\Tencent\\QQ\\Bin\\"));
    try {
        process = pb.start();
    } catch (IOException e) {
        e.printStackTrace();
    }

    }

The result of above is: "Cannot run program "QQ.exe" (in directory "D:\Program Files (x86)\Tencent\QQ\Bin"): CreateProcess error=2, The system cannot find the file specified"

So what does the function of Process.dir()? Somebody told me that the dir i specified is a working directory of running process - it doesn't help to find executable. but the follow code could run rightly

@org.junit.Test
public void test() {
    Process process = null;
    ProcessBuilder pb = new ProcessBuilder("cmd","/c","QQ.exe");
    pb.directory(new File("D:\\Program Files (x86)\\Tencent\\QQ\\Bin\\"));
    try {
        process = pb.start();
    } catch (IOException e) {
        e.printStackTrace();
    }

    }

The qq.exe is not in path。Who can tell me why? I am a chinese. I am not good at english, so Please excuse this wretched apology for my english.

Answer 1

ProcessBuilder.directory(java.io.File) define the working directory of the process, not the "launch", so:

case 1: QQ.exe is started with working directory D:\\Program Files (x86)\\Tencent\\QQ\\Bin\\ but QQ.exe is not found (not in %PATH%)

case 2: cmd.exe is started with working directory D:\\Program Files (x86)\\Tencent\\QQ\\Bin\\ then QQ.exe is launched and found (cmd is in %PATH% and QQ.exe in current working dir)

We can assume that ProcessBuilder starts the process then perform a working directory change.