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

2016-09-10 Thread Geert via Digitalmars-d-learn
I'm sorry for the "necro-bumping", but I've not found the solution for this yet, and i'm getting a different error message at the execution moment: "std.stdio.StdioException@/build/ldc/src/ldc/runtime/phobos/std/stdio.d(4066): Bad file descriptor" This is my testing code: module dd_test;

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

2013-10-11 Thread Colin Grogan
On Thursday, 10 October 2013 at 01:24:03 UTC, Jesse Phillips wrote: On Wednesday, 9 October 2013 at 14:54:32 UTC, Colin Grogan wrote: is blocking. However, its not meant to be blocking is it not? That new /bin/bash process is meant to run in parallel to the main process? I'm not sure exactly

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

2013-10-09 Thread Jesse Phillips
On Wednesday, 9 October 2013 at 14:54:32 UTC, Colin Grogan wrote: is blocking. However, its not meant to be blocking is it not? That new /bin/bash process is meant to run in parallel to the main process? I'm not sure exactly the implementation. But if you're asking to run bash and then print

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

2013-10-09 Thread Colin Grogan
On Wednesday, 9 October 2013 at 14:27:30 UTC, Jesse Phillips wrote: On Wednesday, 9 October 2013 at 11:22:26 UTC, Colin Grogan wrote: " ~/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 int

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

2013-10-09 Thread Jesse Phillips
On Wednesday, 9 October 2013 at 11:22:26 UTC, Colin Grogan wrote: " ~/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. It is extracting the output of t

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

2013-10-09 Thread 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