82 lines
2.4 KiB
C++
82 lines
2.4 KiB
C++
/*
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Copyright (c) 2013 Microsoft Corporation. All rights reserved.
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Released under Apache 2.0 license as described in the file LICENSE.
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Author: Leonardo de Moura
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*/
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#pragma once
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#include <memory>
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#include <utility>
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#include "kernel/environment.h"
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#include "kernel/formatter.h"
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namespace lean {
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class frontend;
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/**
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\brief Expression elaborator for the Lean default frontend, it is responsible for "filling" holes
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in terms left by the user. This is the main module resposible for computing
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the value of implicit arguments. It is based on the general purpose elaborator defined at
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library/elaborator/elaborator.h
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*/
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class frontend_elaborator {
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class imp;
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std::shared_ptr<imp> m_ptr;
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static void print(imp * ptr);
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public:
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explicit frontend_elaborator(frontend const & fe);
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~frontend_elaborator();
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/**
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\brief Elaborate the given expression.
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*/
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expr operator()(expr const & e);
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/**
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\brief Elaborate the given type and expression. The typeof(e) == t.
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This information is used by the elaborator. The result is a new
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elaborated type and expression.
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*/
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std::pair<expr, expr> operator()(name const & n, expr const & t, expr const & e);
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/**
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\brief If \c e is an expression instantiated by the elaborator, then it
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returns the expression (the one with "holes") used to generate \c e.
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Otherwise, it just returns \c e.
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*/
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expr const & get_original(expr const & e) const;
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void set_interrupt(bool flag);
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void interrupt() { set_interrupt(true); }
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void reset_interrupt() { set_interrupt(false); }
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void clear();
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environment const & get_environment() const;
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};
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/**
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\brief Create a choice expression for the given functions.
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It is used to mark which functions can be used in a particular application.
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The elaborator decides which one should be used based on the type of the arguments.
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\pre num_fs >= 2
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*/
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expr mk_choice(unsigned num_fs, expr const * fs);
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/**
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\brief Return true iff \c e is an expression created using \c mk_choice.
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*/
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bool is_choice(expr const & e);
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/**
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\brief Return the number of alternatives in a choice expression.
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We have that <tt>get_num_choices(mk_choice(n, fs)) == n</tt>.
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\pre is_choice(e)
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*/
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unsigned get_num_choices(expr const & e);
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/**
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\brief Return the (i+1)-th alternative of a choice expression.
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\pre is_choice(e)
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\pre i < get_num_choices(e)
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*/
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expr const & get_choice(expr const & e, unsigned i);
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}
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