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