Python regex and string replacement

Question

I'm trying to convert strings in Python such as:

string = 'void demofun(double* output, double db4nsfy[], double VdSGV[], int length)'

into

wrapper = 'void demofun(ref double output, double[] db4nsfy, double[] VdSGV, int length)'

Now for most cases, I'm able to use a trivial combination of while, string.find() and string.replace() to do this because I don't need to meddle with the variable names (such as output or length), but what I can't figure out is replacing these strings:

double db4nsfy[] --> double[] db4nsfy

double[] VdSGV --> double[] VdSGV

How should I do this? I know I will find my answer with some RTFM of regex in Python, but I'm hoping to start with a practical example.

I'm probably wrong - but aren't their better ways instead of manipulating the source code of co-oping between C/C++ and C# ? — Jon Clements
– Jon Clements, Commented Mar 16, 2013 at 5:45
Hm, it'd be annoying if a tool like this already existed but I'd be even happier if it did. I have a lot of Fortran/C numerical routines that I need to interop and they get thrown into many separate .cs source files because of the various things I need done (e.g. testing their speeds against my pure C# implementations). I figured the usage was too specialized but simple enough that a proper tool didn't exist. Had to come up with my own pariah method of indicating which arguments to marshal with the [In] and [Out] attributes. — elleciel
– elleciel, Commented Mar 16, 2013 at 6:07

Braiam · Accepted Answer · 2015-04-05 19:34:56Z

2

You could use re.sub:

>>> import re
>>> re.sub(r'(\w+) (\w+)\[\]', r'\1[] \2', string)
    'void demofun(double* output, double[] db4nsfy, double[] VdSGV, int length)'

(\w+) (\w+)\[\] matches two "words" wrapped in capturing groups and brackets.
\1 and \2 refer to the stuff captured by those groups.

edited Apr 5, 2015 at 19:34

Braiam

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answered Mar 16, 2013 at 5:43

Blender

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3 Comments

elleciel Over a year ago

Awesome! Explains the solution I needed. Could you suggest the benefits that I'd gain from using something like AStyle in this particular instance? Be brutally honest ;)

Blender Over a year ago

@ephedyn: Probably ease of use and versatility. Regex is clunky and wasn't made specifically to beautify source files.

elleciel Over a year ago

I see. I'm actually using Python to generate C# wrappers for a bunch of Fortran/C subroutines. I'll pick up a code formatter and try it out on the routines themselves, thanks a lot!

thkang · Accepted Answer · 2013-03-16 05:50:38Z

verbose, but without regex and handles both pointer and arrays(and also without regex):

def new_arguments(func_string):
    def argument_replace(arguments):
        new_arguments = []
        for argument in arguments.split(', '):
            typ, var = argument.split()
            if typ.endswith('*'):
                typ = 'ref ' + typ.replace('*', '')
            if var.endswith('[]'):
                var = var.replace('[]', '')
                typ += '[]'
            new_arguments.append(' '.join([typ, var]))
        return ', '.join(new_arguments)

    func_name = func_string[:func_string.index('(')]
    arguments = func_string[func_string.index('(')+1:func_string.index(')')]

    return ''.join((func_name, '(', argument_replace(arguments), ')'))

string = 'void demofun(double* output, double db4nsfy[], double VdSGV[], int length)'
print new_arguments(string)
#void demofun(ref double output, double[] db4nsfy, double[] VdSGV, int length)

fahrbach · Accepted Answer · 2013-03-16 05:52:50Z

1

This is an intuitive approach without regex.

s = 'void demofun(double* output, double db4nsfy[], double VdSGV[], int length)'
s = s.split()
for i in range(len(s)):
    if s[i][-3:] == '[],':
        s[i] = s[i][:-3] + ','
        s[i-1] = s[i-1] + '[]'
    elif s[i][-3:] == '[])':
        s[i] = s[i][:-3] + ')'
        s[i-1] = s[i-1] + '[]'
s = ' '.join(s)
print s
# void demofun(double* output, double[] db4nsfy, double[] VdSGV, int length)

answered Mar 16, 2013 at 5:52

fahrbach

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