mirror of https://github.com/python/cpython
104 lines
3.9 KiB
TeX
104 lines
3.9 KiB
TeX
\section{\module{codeop} ---
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Compile Python code}
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% LaTeXed from excellent doc-string.
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\declaremodule{standard}{codeop}
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\sectionauthor{Moshe Zadka}{moshez@zadka.site.co.il}
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\sectionauthor{Michael Hudson}{mwh@python.net}
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\modulesynopsis{Compile (possibly incomplete) Python code.}
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The \module{codeop} module provides utilities upon which the Python
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read-eval-print loop can be emulated, as is done in the
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\refmodule{code} module. As a result, you probably don't want to use
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the module directly; if you want to include such a loop in your
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program you probably want to use the \refmodule{code} module instead.
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There are two parts to this job:
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\begin{enumerate}
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\item Being able to tell if a line of input completes a Python
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statement: in short, telling whether to print
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`\code{>\code{>}>~}' or `\code{...~}' next.
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\item Remembering which future statements the user has entered, so
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subsequent input can be compiled with these in effect.
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\end{enumerate}
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The \module{codeop} module provides a way of doing each of these
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things, and a way of doing them both.
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To do just the former:
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\begin{funcdesc}{compile_command}
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{source\optional{, filename\optional{, symbol}}}
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Tries to compile \var{source}, which should be a string of Python
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code and return a code object if \var{source} is valid
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Python code. In that case, the filename attribute of the code object
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will be \var{filename}, which defaults to \code{'<input>'}.
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Returns \code{None} if \var{source} is \emph{not} valid Python
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code, but is a prefix of valid Python code.
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If there is a problem with \var{source}, an exception will be raised.
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\exception{SyntaxError} is raised if there is invalid Python syntax,
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and \exception{OverflowError} or \exception{ValueError} if there is an
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invalid literal.
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The \var{symbol} argument determines whether \var{source} is compiled
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as a statement (\code{'single'}, the default) or as an expression
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(\code{'eval'}). Any other value will cause \exception{ValueError} to
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be raised.
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\strong{Caveat:}
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It is possible (but not likely) that the parser stops parsing
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with a successful outcome before reaching the end of the source;
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in this case, trailing symbols may be ignored instead of causing an
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error. For example, a backslash followed by two newlines may be
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followed by arbitrary garbage. This will be fixed once the API
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for the parser is better.
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\end{funcdesc}
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\begin{classdesc}{Compile}{}
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Instances of this class have \method{__call__()} methods indentical in
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signature to the built-in function \function{compile()}, but with the
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difference that if the instance compiles program text containing a
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\module{__future__} statement, the instance 'remembers' and compiles
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all subsequent program texts with the statement in force.
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\end{classdesc}
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\begin{classdesc}{CommandCompiler}{}
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Instances of this class have \method{__call__()} methods identical in
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signature to \function{compile_command()}; the difference is that if
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the instance compiles program text containing a \code{__future__}
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statement, the instance 'remembers' and compiles all subsequent
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program texts with the statement in force.
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\end{classdesc}
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A note on version compatibility: the \class{Compile} and
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\class{CommandCompiler} are new in Python 2.2. If you want to enable
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the future-tracking features of 2.2 but also retain compatibility with
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2.1 and earlier versions of Python you can either write
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\begin{verbatim}
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try:
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from codeop import CommandCompiler
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compile_command = CommandCompiler()
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del CommandCompiler
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except ImportError:
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from codeop import compile_command
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\end{verbatim}
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which is a low-impact change, but introduces possibly unwanted global
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state into your program, or you can write:
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\begin{verbatim}
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try:
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from codeop import CommandCompiler
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except ImportError:
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def CommandCompiler():
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from codeop import compile_command
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return compile_command
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\end{verbatim}
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and then call \code{CommandCompiler} every time you need a fresh
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compiler object.
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