std.process spawnShell/pipeShell dont capture output of the shell

[Colin Grogan]

Hi folks,

Is there anyway to make std.process.spawnShell or 
std.process.pipeShell capture the output of the shell interface?

For example, the code:

import std.stdio;
import std.process;
void main(string[] args){

    writefln("Executing: %s", args[1]);
    auto processPipes = pipeShell(args[1], Redirect.all);
                
    foreach(str; processPipes.stdout.byLine){
        writefln("STDOUT: %s",str);
    }
        
    foreach(str; processPipes.stderr.byLine){
        writefln("STDERR: %s",str);
    }   
}

is compiled to an executable called "mess".
When executed with the following produces:
"
~/test$ ./mess ls
Executing: ls
STDOUT: mess
STDOUT: text.txt

"
Thats all fine, however, I'd expect it to print another "~/test$" 
at the end, as if its an interactive shell waiting for input.

Is that an incorrect assumption to make or am I doing something 
wrong that is stopping that from happening?


Also, If I run /bin/bash through program, it hangs.

"
~/test$ ./mess /bin/bash
Executing: /bin/bash

"

Thanks,
Colin