How to let BulletClient not print "argv[0]=" at initialization

[mengyuest]

I have pybullet==3.2.5, and when I try to create a PyBullet client connection via bullet_client.BulletClient(connection_mode=p.DIRECT), at the very beginning, it will output "argv[0]=". This could be not very pleasant if I want to build many connections. A minimal example is the following:

import pybullet as p
from pybullet_utils import bullet_client as bc
for _ in range(10):
    client = bc.BulletClient(connection_mode=p.DIRECT)

This will output

argv[0]=
argv[0]=
argv[0]=
argv[0]=
argv[0]=
argv[0]=
argv[0]=
argv[0]=
argv[0]=
argv[0]=

Any idea to solve this? Thanks.

[EtorArza]

Hey, I also encountered this problem and used this workaround, although its a dirty hack :(

I was unable to find a simpler way to do this.

import ctypes
import sys
class RedirectStream(object):

  @staticmethod
  def _flush_c_stream(stream):
    streamname = stream.name[1:-1]
    libc = ctypes.CDLL(None)
    libc.fflush(ctypes.c_void_p.in_dll(libc, streamname))

  def __init__(self, stream=sys.stdout, file=os.devnull):
    self.stream = stream
    self.file = file

  def __enter__(self):
    self.stream.flush()  # ensures python stream unaffected 
    self.fd = open(self.file, "w+")
    self.dup_stream = os.dup(self.stream.fileno())
    os.dup2(self.fd.fileno(), self.stream.fileno()) # replaces stream
  
  def __exit__(self, type, value, traceback):
    RedirectStream._flush_c_stream(self.stream)  # ensures C stream buffer empty
    os.dup2(self.dup_stream, self.stream.fileno()) # restores stream
    os.close(self.dup_stream)
    self.fd.close()

And then run the code that generates argv[0]= inside a with RedirectStream(sys.stdout):. So in your case,

with RedirectStream(sys.stdout):
    for _ in range(10):
        client = bc.BulletClient(connection_mode=p.DIRECT)

If you find a simpler way let me know :)