Windows2003-3790/multimedia/directx/dmusic/dmscript/engexpr.cpp
2020-09-30 16:53:55 +02:00

165 lines
4.2 KiB
C++

// Copyright (c) 1999 Microsoft Corporation. All rights reserved.
//
// Implementation of ExprBlock.
//
#include "stdinc.h"
#include "engexpr.h"
HRESULT
Expression::Generate()
{
HRESULT hr = InfixToPostfix();
if (FAILED(hr))
{
// clean up the working stack
while (!m_stack.empty())
m_stack.pop();
return hr;
}
return S_OK;
}
int Precedence(Token t)
{
switch(t)
{
case TOKEN_op_pow: return 10;
case TOKEN_sub: return 9; // unary - (negation)
case TOKEN_op_mult: return 8;
case TOKEN_op_div: return 7;
case TOKEN_op_mod: return 6;
case TOKEN_op_plus:
case TOKEN_op_minus: return 5;
case TOKEN_op_lt:
case TOKEN_op_leq:
case TOKEN_op_gt:
case TOKEN_op_geq:
case TOKEN_op_eq:
case TOKEN_op_neq:
case TOKEN_is: return 4;
case TOKEN_op_not: return 3;
case TOKEN_and: return 2;
case TOKEN_or: return 1;
case TOKEN_lparen: return 0;
default:
assert(false);
return 12;
}
}
// Infix to postfix conversion is performed by a single scan of the infix expression blocks.
// A stack is used to hold some of the blocks before they eventually are appended to the postfix expression.
//
// The algorithm follows the following rules:
// * If the current item is an value it is immediately appended.
// * If the current item is an operator, pop and append each operator on the stack until one is encountered that:
// - has lower precedence than the current operator OR
// - is a left paren OR
// - is a unary operator and the current item is also unary
// Once done with this popping, push the current item onto the stack.
// * If the current item is a left paren, push it onto the stack.
// * If the current item is a right paren, pop and append all the operators until the matching left paren is found.
// Discard the left and right paren as parens are not needed in postfix.
// * After scanning all items of the input, pop and append any operators that remain on the stack.
//
// Before working with this code, try out a few expressions on paper to see how this works.
HRESULT
Expression::InfixToPostfix()
{
assert(m_stack.empty());
HRESULT hr = S_OK;
ExprBlocks::index iLast = m_e.Next();
assert(iLast > 0);
for (ExprBlocks::index i = 0; i < iLast; ++i)
{
const ExprBlock &b = m_e[i];
if (b.k == ExprBlock::_val || b.k == ExprBlock::_call)
{
// this is an operand -- send it directly to the postfix output
hr = m_eblocks.Add(b);
if (FAILED(hr))
return hr;
}
else
{
if (b.op == TOKEN_rparen)
{
// pop whatever's left until the matching lparen
for (;;)
{
if (m_stack.empty())
{
assert(false);
return E_FAIL;
}
ExprBlock bPop = m_stack.top();
if (bPop.op == TOKEN_lparen)
{
m_stack.pop();
break;
}
hr = m_eblocks.Add(bPop);
if (FAILED(hr))
return hr;
m_stack.pop();
}
continue;
}
else if (b.op != TOKEN_lparen)
{
// Pop all operators of lower precedence off the stack. (This won't pass a left paren because its precedence is set to 0.)
// Exception: don't pop a unary operator if the new one is also unary.
int iNewPrecidence = Precedence(b.op);
bool fNewUnary = b.op == TOKEN_sub || b.op == TOKEN_op_not;
while (!m_stack.empty()) // note that there's a break inside the loop as well
{
ExprBlock bPop = m_stack.top();
if (Precedence(bPop.op) < iNewPrecidence)
break;
if (fNewUnary && (bPop.op == TOKEN_sub || bPop.op == TOKEN_op_not))
break;
hr = m_eblocks.Add(bPop);
if (FAILED(hr))
return hr;
m_stack.pop();
}
}
// now push the new operator onto the stack
hr = m_stack.push(b);
if (FAILED(hr))
return hr;
}
}
while (!m_stack.empty())
{
ExprBlock bPop = m_stack.top();
if (bPop.op == TOKEN_lparen)
{
assert(false);
return E_FAIL;
}
hr = m_eblocks.Add(bPop);
if (FAILED(hr))
return hr;
m_stack.pop();
}
// Add the teminating (_end) block
hr = m_eblocks.Add(ExprBlock(ExprBlock::cons_end()));
if (FAILED(hr))
return hr;
return S_OK;
}