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How can I parse a C format string in Python?

I have this code in my C file:

printf("Worker name is %s and id is %d", worker.name, worker.id);

I want, with Python, to be able to parse the format string and locate the "%s" and "%d".

So I want to have a function:

>>> my_function("Worker name is %s and id is %d")
[Out1]: ((15, "%s"), (28, "%d))

I've tried to achieve this using libclang's Python bindings, and with pycparser, but I didn't see how can this be done with these tools.

I've also tried using regex to solve this, but this is not simple at all - think about use cases when the printf has "%%s" and stuff like this.

Both gcc and clang obviously do this as part of compiling - have no one exported this logic to Python?

like image 583
speller Avatar asked May 03 '15 07:05

speller


2 Answers

You can certainly find properly formatted candidates with a regex.

Take a look at the definition of the C Format Specification. (Using Microsofts, but use what you want.)

It is:

%[flags] [width] [.precision] [{h | l | ll | w | I | I32 | I64}] type

You also have the special case of %% which becomes % in printf.

You can translate that pattern into a regex:

(                                 # start of capture group 1
%                                 # literal "%"
(?:                               # first option
(?:[-+0 #]{0,5})                  # optional flags
(?:\d+|\*)?                       # width
(?:\.(?:\d+|\*))?                 # precision
(?:h|l|ll|w|I|I32|I64)?           # size
[cCdiouxXeEfgGaAnpsSZ]            # type
) |                               # OR
%%)                               # literal "%%"

Demo

And then into a Python regex:

import re

lines='''\
Worker name is %s and id is %d
That is %i%%
%c
Decimal: %d  Justified: %.6d
%10c%5hc%5C%5lc
The temp is %.*f
%ss%lii
%*.*s | %.3d | %lC | %s%%%02d'''

cfmt='''\
(                                  # start of capture group 1
%                                  # literal "%"
(?:                                # first option
(?:[-+0 #]{0,5})                   # optional flags
(?:\d+|\*)?                        # width
(?:\.(?:\d+|\*))?                  # precision
(?:h|l|ll|w|I|I32|I64)?            # size
[cCdiouxXeEfgGaAnpsSZ]             # type
) |                                # OR
%%)                                # literal "%%"
'''

for line in lines.splitlines():
    print '"{}"\n\t{}\n'.format(line, 
           tuple((m.start(1), m.group(1)) for m in re.finditer(cfmt, line, flags=re.X))) 

Prints:

"Worker name is %s and id is %d"
    ((15, '%s'), (28, '%d'))

"That is %i%%"
    ((8, '%i'), (10, '%%'))

"%c"
    ((0, '%c'),)

"Decimal: %d  Justified: %.6d"
    ((9, '%d'), (24, '%.6d'))

"%10c%5hc%5C%5lc"
    ((0, '%10c'), (4, '%5hc'), (8, '%5C'), (11, '%5lc'))

"The temp is %.*f"
    ((12, '%.*f'),)

"%ss%lii"
    ((0, '%s'), (3, '%li'))

"%*.*s | %.3d | %lC | %s%%%02d"
    ((0, '%*.*s'), (8, '%.3d'), (15, '%lC'), (21, '%s'), (23, '%%'), (25, '%02d'))
like image 170
dawg Avatar answered Oct 24 '22 10:10

dawg


A simple implementation might be the following generator:

def find_format_specifiers(s):
    last_percent = False
    for i in range(len(s)):
        if s[i] == "%" and not last_percent:
            if s[i+1] != "%":
                yield (i, s[i:i+2])
            last_percent = True
        else:
            last_percent = False

>>> list(find_format_specifiers("Worker name is %s and id is %d but %%q"))
[(15, '%s'), (28, '%d')]

This can be fairly easily extended to handle additional format specifier information like width and precision, if needed.

like image 23
Greg Hewgill Avatar answered Oct 24 '22 12:10

Greg Hewgill