On Error Resume Next in Python

Question

Snippet 1

do_magic() # Throws exception, doesn't execute do_foo and do_bar
do_foo()
do_bar()

Snippet 2

try:
    do_magic() # Doesn't throw exception, doesn't execute do_foo and do_bar
    do_foo() 
    do_bar()
except:
    pass

Snippet 3

try: do_magic(); except: pass
try: do_foo()  ; except: pass
try: do_bar()  ; except: pass

Is there a way to write code snippet 3 elegantly?

• if do_magic() fails or not, do_foo() and do_bar() should be executed.
• if do_foo() fails or not, do_bar() should be executed.

In Basic/Visual Basic/VBS, there's a statement called On Error Resume Next which does this.

Answer 1

In Python 3.4 onwards, you can use contextlib.suppress:

from contextlib import suppress

with suppress(Exception): # or, better, a more specific error (or errors)
    do_magic()
with suppress(Exception):
    do_foo()
with suppress(Exception):
    do_bar()

Alternatively, fuckit.

Answer 2

If all three functions accept same number of parameters:

for f in (do_magic, do_foo, do_bar):
    try:
        f()
    except:
        pass

Otherwise, wrap the function call with lambda.

for f in (do_magic, lambda: do_foo(arg1, arg2)):
    try:
        f()
    except:
        pass

Answer 3

If there are no parameters...

funcs = do_magic, do_foo, do_bar

for func in funcs:
    try:
        func()
    except:
        continue

Answer 4

If you are the one coding the fucntions, why not program the functions to return status codes? Then they will be atomic and you wont have to capture the error in the main section. You will also be able to perform roll back or alternate coding on failure.

def do_magic():
    try:
        #do something here
        return 1
    except:
        return 0

in main program..

if do_magic() = 0:
   #do something useful or not...

if do_foo() = 0:
   #do something useful or not...

if do_bar() = 0:
   #do something useful or not...

Answer 5

A lot of ident, but it works

try:
    do_magic()
finally:
    try:
        do_foo()
    finally:
        try:
            do_bar()
        finally:
            pass

Answer 6

you could try a nested ´try´ loop, alltho that might not be as elegantly pythonic as you might want. the ´lambda´ solution is is also a good way to go, did not mention because it was done in the previous answer

edit:

try:
    do_magic()
finally:
    try:
        do_foo()
    finally:
        try:
            do_bar()
        except:
            pass

edit 2:

well damnnit, this answer just got posted seconds beforehand again :|

Answer 7

In the question, Snippet 3 does not work but will work if you don't mind splitting each line over two lines...

try: do_magic()
except: pass
try: do_foo()
except: pass
try: do_bar()
except: pass

A working example..

import sys
a1 = "No_Arg1"
a2 = "No_Arg2"
a3 = "No_Arg3"
try: a1 = sys.argv[1]
except: pass
try: a2 = sys.argv[2]
except: pass
try: a3 = sys.argv[3]
except: pass

print a1, a2, a3

..if you save this to test.py and then at a CMD prompt in windows simply type test.py it will return No_Arg1 No_Arg2 No_Arg3 because there were no arguments. However, if you supply some arguments, if type test.py 111 222 it will return 111 222 No_Arg3 etc. (Tested - Windows 7, python2.7).

IMHO this is far more elegant than the nesting example replies. It also works exactly like On Error Resume Next and I use it when translating from VB6. One issue is that the try lines cannot contain a conditional. I have found that as a rule, python cannot contain more than one : in a line. That said, it simply means splitting the statement over 3 lines etc.