Question

Are Python’s built-in functions not available for use as keyword defaults, or should I be using some other way of referring to a function?

I wanted to write a function like this:

def isPNGBlock(bytes, blockLen, pngOffset=0, pngErrorLogger=print):
    ...
    try:
        r.validate_signature()
        width, height, pixels, metadata = r.read(lenient=True)
    except png.Error as e:
        pngErrorLogger(e)

Instead I’ve had to settle for doing this with a default argument of None as a flag value.

def isPNGBlock(bytes, blockLen, pngOffset=0, pngErrorLogger=None):
    ...
    try:
        r.validate_signature()
        width, height, pixels, metadata = r.read(lenient=True)
    except png.Error as e:
        if pngErrorLogger is None:
            print(e)
        else:
            pngErrorLogger(e)

or using a wrapper function:

def defaultLogger(str):
    print(str)

def isPNGBlock(bytes, blockLen, pngOffset=0, pngErrorLogger=defaultLogger ):
    ...
    try:
        r.validate_signature()
        width, height, pixels, metadata = r.read(lenient=True)
    except png.Error as e:
        pngErrorLogger(e)

Answer 1

Are Python’s built-in functions not available for use as keyword defaults

They are available to use just like any other function.

However, in Python 2 print is a statement, not a function. It became a function in Python 3, so your code will work there. It will also work in recent versions of Python 2 if you use from __future__ import print_function. For example, using Python 2.7.3:

In [2]: from __future__ import print_function

In [3]: def isPNGBlock(bytes, blockLen, pngOffset=0, pngErrorLogger=print):
   ...:     pngErrorLogger('test')
   ...:     

In [4]: isPNGBlock(0, 0)
test

If you can’t use print as a function, you could either write a wrapper, or use sys.stdout.write:

In [7]: isPNGBlock(0, 0, 0, sys.stdout.write)
test