PyCharm debugger : unable to step in socket module
PyCharm debugger does not step in when using a simple socket object.
from socket import socket
s = socket()
s.connect(('localhost', 50000))
s.close()
This simple code runs fine, but debugger steps over every instruction even when trying to step in (F7).
I know socket is just an interface, yet when i display implementation, ssl.py shows up, so PyCharm does know where to find it.
Is it normal behavior? I have the feeling I am missing something.
How can I step into the executing code?
P.S. : Option "Do not step into library scripts" is disabled
System info:
Mint 18.3
Python 3.5
PyCharm 2018.2 (Edu)
Build #PE-182.3684.126, built on July 30, 2018
JRE: 1.8.0_152-release-1248-b8 amd64
JVM: OpenJDK 64-Bit Server VM by JetBrains s.r.o
Linux 4.13.0-39-generic
python debugging pycharm
asked Nov 20 at 12:29
Bastien
4281515
add a comment |
PyCharm debugger does not step in when using a simple socket object.
from socket import socket
s = socket()
s.connect(('localhost', 50000))
s.close()
This simple code runs fine, but debugger steps over every instruction even when trying to step in (F7).
I know socket is just an interface, yet when i display implementation, ssl.py shows up, so PyCharm does know where to find it.
Is it normal behavior? I have the feeling I am missing something.
How can I step into the executing code?
P.S. : Option "Do not step into library scripts" is disabled
System info:
Mint 18.3
Python 3.5
PyCharm 2018.2 (Edu)
Build #PE-182.3684.126, built on July 30, 2018
JRE: 1.8.0_152-release-1248-b8 amd64
JVM: OpenJDK 64-Bit Server VM by JetBrains s.r.o
Linux 4.13.0-39-generic
python debugging pycharm
asked Nov 20 at 12:29
Bastien
4281515
add a comment |
PyCharm debugger does not step in when using a simple socket object.
from socket import socket
s = socket()
s.connect(('localhost', 50000))
s.close()
This simple code runs fine, but debugger steps over every instruction even when trying to step in (F7).
I know socket is just an interface, yet when i display implementation, ssl.py shows up, so PyCharm does know where to find it.
Is it normal behavior? I have the feeling I am missing something.
How can I step into the executing code?
P.S. : Option "Do not step into library scripts" is disabled
System info:
Mint 18.3
Python 3.5
PyCharm 2018.2 (Edu)
Build #PE-182.3684.126, built on July 30, 2018
JRE: 1.8.0_152-release-1248-b8 amd64
JVM: OpenJDK 64-Bit Server VM by JetBrains s.r.o
Linux 4.13.0-39-generic
python debugging pycharm
asked Nov 20 at 12:29
Bastien
4281515
PyCharm debugger does not step in when using a simple socket object.
from socket import socket
s = socket()
s.connect(('localhost', 50000))
s.close()
This simple code runs fine, but debugger steps over every instruction even when trying to step in (F7).
I know socket is just an interface, yet when i display implementation, ssl.py shows up, so PyCharm does know where to find it.
Is it normal behavior? I have the feeling I am missing something.
How can I step into the executing code?
P.S. : Option "Do not step into library scripts" is disabled
System info:
Mint 18.3
Python 3.5
PyCharm 2018.2 (Edu)
Build #PE-182.3684.126, built on July 30, 2018
JRE: 1.8.0_152-release-1248-b8 amd64
JVM: OpenJDK 64-Bit Server VM by JetBrains s.r.o
Linux 4.13.0-39-generic
python debugging pycharm
python debugging pycharm
asked Nov 20 at 12:29
Bastien
4281515
asked Nov 20 at 12:29
Bastien
4281515
asked Nov 20 at 12:29
Bastien
4281515
asked Nov 20 at 12:29
Bastien
4281515
asked Nov 20 at 12:29
Bastien
4281515
4281515
add a comment |
add a comment |
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