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Highlights: import and friends will understand any of \r, \n and \r\n as end of line. Python file input will do the same if you use mode 'U'. Everything can be disabled by configuring with --without-universal-newlines. See PEP278 for details.
83 lines
2.9 KiB
Python
83 lines
2.9 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 imp
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MAGIC = imp.get_magic()
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__all__ = ["compile"]
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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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import os, marshal, __builtin__
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f = open(file, 'U')
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try:
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timestamp = long(os.fstat(f.fileno())[8])
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except AttributeError:
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timestamp = long(os.stat(file)[8])
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codestring = f.read()
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# If parsing from a string, line breaks are \n (see parsetok.c:tok_nextc)
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# Replace will return original string if pattern is not found, so
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# we don't need to check whether it is found first.
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codestring = codestring.replace("\r\n","\n")
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codestring = codestring.replace("\r","\n")
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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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import traceback, sys
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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 not cfile:
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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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if os.name == 'mac':
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import macfs
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macfs.FSSpec(cfile).SetCreatorType('Pyth', 'PYC ')
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