Why does Process.Start(ProcessStartInfo) fail?

Question

We've developed a new WPF application, and I have had difficulty launching it from external C# script. While calling `Process.Start(ProcessStartInfo)` method with `ProcessStartInfo` object which is initialized with both `WorkingDirectory` and `FileName` successes, init the `FileName` property only fails to launch. This is not the case when calling any other application. My question - Does different approach to start process has different logic? See code for more details: ``` public void LaunchApp(){ /********************************/ /* This code PASSES */ /********************************/ var pStartInfoCalc1 = new ProcessStartInfo { FileName = @"C:\Windows\system32\calc.exe", }; Process.Start(pStartInfoCalc1); /*****************************/ /* !!!This code FAILS !!! */ /*****************************/ var pStartInfo1 = new ProcessStartInfo { FileName = @"C:\Program Files\MyAppFolder\MyApp.exe", }; Process.Start(pStartInfo1); /********************************/ /* This code PASSES */…

Accepted Answer

When you do not specify the working directory, the new process will inherit the working directory of your process. That is, the new process will inherit the working directory of the process that called `Process.Start()`.

That's the only difference between the two attempts to start `MyApp`. One of them inherits the working directory, and one of them specifies it. Clearly `MyApp` does not like to run with the initial working directory being the directory of your parent process.

Why that is so, I cannot say for sure. It would seem that `MyApp` is attempting some XML parsing at startup. So perhaps that XML parsing reads a file that is presumed to be in the working directory. But in fact the file is located in the same directory as the executable.

If that is the case then you need to modify `MyApp` to solve the problem. Instead of using a relative path for this XML file, you need to build an absolute path based on the directory of the executable file.

The `MyApp` startup code can that directory, like this:

```
string ExeDir = System.IO.Path.GetDirectoryName(System.Reflection.Assembly.
    GetExecutingAssembly().Location));
```

And then you would use `Path.Combine` to form the full path to the XML file.

Why does Process.Start(ProcessStartInfo) fail?

Solution

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