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107 lines
3.3 KiB
Python
107 lines
3.3 KiB
Python
"""Routine to "compile" a .py file to a .pyc (or .pyo) file.
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This module has intimate knowledge of the format of .pyc files.
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"""
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import __builtin__
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import imp
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import marshal
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import os
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import sys
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import traceback
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MAGIC = imp.get_magic()
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__all__ = ["compile", "main"]
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# Define an internal helper according to the platform
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if os.name == "mac":
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import macfs
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def set_creator_type(file):
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macfs.FSSpec(file).SetCreatorType('Pyth', 'PYC ')
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else:
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def set_creator_type(file):
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pass
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def wr_long(f, x):
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"""Internal; write a 32-bit int to a file in little-endian order."""
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f.write(chr( x & 0xff))
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f.write(chr((x >> 8) & 0xff))
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f.write(chr((x >> 16) & 0xff))
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f.write(chr((x >> 24) & 0xff))
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def compile(file, cfile=None, dfile=None):
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"""Byte-compile one Python source file to Python bytecode.
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Arguments:
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file: source filename
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cfile: target filename; defaults to source with 'c' or 'o' appended
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('c' normally, 'o' in optimizing mode, giving .pyc or .pyo)
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dfile: purported filename; defaults to source (this is the filename
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that will show up in error messages)
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Note that it isn't necessary to byte-compile Python modules for
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execution efficiency -- Python itself byte-compiles a module when
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it is loaded, and if it can, writes out the bytecode to the
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corresponding .pyc (or .pyo) file.
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However, if a Python installation is shared between users, it is a
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good idea to byte-compile all modules upon installation, since
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other users may not be able to write in the source directories,
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and thus they won't be able to write the .pyc/.pyo file, and then
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they would be byte-compiling every module each time it is loaded.
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This can slow down program start-up considerably.
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See compileall.py for a script/module that uses this module to
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byte-compile all installed files (or all files in selected
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directories).
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"""
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f = open(file, 'U')
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try:
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timestamp = long(os.fstat(f.fileno()).st_mtime)
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except AttributeError:
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timestamp = long(os.stat(file).st_mtime)
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codestring = f.read()
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f.close()
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if codestring and codestring[-1] != '\n':
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codestring = codestring + '\n'
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try:
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codeobject = __builtin__.compile(codestring, dfile or file, 'exec')
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except SyntaxError, detail:
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lines = traceback.format_exception_only(SyntaxError, detail)
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for line in lines:
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sys.stderr.write(line.replace('File "<string>"',
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'File "%s"' % (dfile or file)))
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return
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if cfile is None:
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cfile = file + (__debug__ and 'c' or 'o')
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fc = open(cfile, 'wb')
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fc.write('\0\0\0\0')
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wr_long(fc, timestamp)
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marshal.dump(codeobject, fc)
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fc.flush()
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fc.seek(0, 0)
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fc.write(MAGIC)
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fc.close()
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set_creator_type(cfile)
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def main(args=None):
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"""Compile several source files.
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The files named in 'args' (or on the command line, if 'args' is
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not specified) are compiled and the resulting bytecode is cached
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in the normal manner. This function does not search a directory
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structure to locate source files; it only compiles files named
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explicitly.
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"""
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if args is None:
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args = sys.argv[1:]
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for filename in args:
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compile(filename)
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if __name__ == "__main__":
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main()
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