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pytqt/pylupdate3/numberh.cpp

231 lines
6.4 KiB

/**********************************************************************
** Copyright (C) 2000 Trolltech AS. All rights reserved.
**
** numberh.cpp
**
** This file is part of Qt Linguist.
**
** See the file LICENSE included in the distribution for the usage
** and distribution terms.
**
** The file is provided AS IS with NO WARRANTY OF ANY KIND,
** INCLUDING THE WARRANTY OF DESIGN, MERCHANTABILITY AND FITNESS FOR
** A PARTICULAR PURPOSE.
**
**********************************************************************/
#include <qmemarray.h>
#include <qcstring.h>
#include <qmap.h>
#include <qstringlist.h>
#include <ctype.h>
#include <metatranslator.h>
typedef QMap<QCString, MetaTranslatorMessage> TMM;
typedef QValueList<MetaTranslatorMessage> TML;
static bool isDigitFriendly( int c )
{
return ispunct( c ) || isspace( c );
}
static int numberLength( const char *s )
{
int i = 0;
if ( isdigit(s[0]) ) {
do {
i++;
} while ( isdigit(s[i]) ||
(isDigitFriendly(s[i]) &&
(isdigit(s[i + 1]) ||
(isDigitFriendly(s[i + 1]) && isdigit(s[i + 2])))) );
}
return i;
}
/*
Returns a version of 'key' where all numbers have been replaced by zeroes. If
there were none, returns "".
*/
static QCString zeroKey( const char *key )
{
QCString zeroed( strlen(key) + 1 );
char *z = zeroed.data();
int i = 0, j = 0;
int len;
bool metSomething = FALSE;
while ( key[i] != '\0' ) {
len = numberLength( key + i );
if ( len > 0 ) {
i += len;
z[j++] = '0';
metSomething = TRUE;
} else {
z[j++] = key[i++];
}
}
z[j] = '\0';
if ( metSomething )
return zeroed;
else
return "";
}
static QString translationAttempt( const QString& oldTranslation,
const char *oldSource,
const char *newSource )
{
int p = zeroKey( oldSource ).contains( '0' );
int oldSourceLen = qstrlen( oldSource );
QString attempt;
QStringList oldNumbers;
QStringList newNumbers;
QMemArray<bool> met( p );
QMemArray<int> matchedYet( p );
int i, j;
int k = 0, ell, best;
int m, n;
int pass;
/*
This algorithm is hard to follow, so we'll consider an example
all along: oldTranslation is "XeT 3.0", oldSource is "TeX 3.0"
and newSource is "XeT 3.1".
First, we set up two tables: oldNumbers and newNumbers. In our
example, oldNumber[0] is "3.0" and newNumber[0] is "3.1".
*/
for ( i = 0, j = 0; i < oldSourceLen; i++, j++ ) {
m = numberLength( oldSource + i );
n = numberLength( newSource + j );
if ( m > 0 ) {
oldNumbers.append( QCString(oldSource + i, m + 1) );
newNumbers.append( QCString(newSource + j, n + 1) );
i += m;
j += n;
met[k] = FALSE;
matchedYet[k] = 0;
k++;
}
}
/*
We now go over the old translation, "XeT 3.0", one letter at a
time, looking for numbers found in oldNumbers. Whenever such a
number is met, it is replaced with its newNumber equivalent. In
our example, the "3.0" of "XeT 3.0" becomes "3.1".
*/
for ( i = 0; i < (int) oldTranslation.length(); i++ ) {
attempt += oldTranslation[i];
for ( k = 0; k < p; k++ ) {
if ( oldTranslation[i] == oldNumbers[k][matchedYet[k]] )
matchedYet[k]++;
else
matchedYet[k] = 0;
}
/*
Let's find out if the last character ended a match. We make
two passes over the data. In the first pass, we try to
match only numbers that weren't matched yet; if that fails,
the second pass does the trick. This is useful in some
suspicious cases, flagged below.
*/
for ( pass = 0; pass < 2; pass++ ) {
best = p; // an impossible value
for ( k = 0; k < p; k++ ) {
if ( (!met[k] || pass > 0) &&
matchedYet[k] == (int) oldNumbers[k].length() &&
numberLength(oldTranslation.latin1() + (i + 1) -
matchedYet[k]) == matchedYet[k] ) {
// the longer the better
if ( best == p || matchedYet[k] > matchedYet[best] )
best = k;
}
}
if ( best != p ) {
attempt.truncate( attempt.length() - matchedYet[best] );
attempt += newNumbers[best];
met[best] = TRUE;
for ( k = 0; k < p; k++ )
matchedYet[k] = 0;
break;
}
}
}
/*
We flag two kinds of suspicious cases. They are identified as
such with comments such as "{2000?}" at the end.
Example of the first kind: old source text "TeX 3.0" translated
as "XeT 2.0" is flagged "TeX 2.0 {3.0?}", no matter what the
new text is.
*/
for ( k = 0; k < p; k++ ) {
if ( !met[k] )
attempt += QString( " {" ) + newNumbers[k] + QString( "?}" );
}
/*
Example of the second kind: "1 of 1" translated as "1 af 1",
with new source text "1 of 2", generates "1 af 2 {1 or 2?}"
because it's not clear which of "1 af 2" and "2 af 1" is right.
*/
for ( k = 0; k < p; k++ ) {
for ( ell = 0; ell < p; ell++ ) {
if ( k != ell && oldNumbers[k] == oldNumbers[ell] &&
newNumbers[k] < newNumbers[ell] )
attempt += QString( " {" ) + newNumbers[k] + QString( " or " ) +
newNumbers[ell] + QString( "?}" );
}
}
return attempt;
}
/*
Augments a MetaTranslator with translations easily derived from
similar existing (probably obsolete) translations.
For example, if "TeX 3.0" is translated as "XeT 3.0" and "TeX 3.1"
has no translation, "XeT 3.1" is added to the translator and is
marked Unfinished.
*/
void applyNumberHeuristic( MetaTranslator *tor, bool verbose )
{
TMM translated, untranslated;
TMM::Iterator t, u;
TML all = tor->messages();
TML::Iterator it;
int inserted = 0;
for ( it = all.begin(); it != all.end(); ++it ) {
if ( (*it).type() == MetaTranslatorMessage::Unfinished ) {
if ( (*it).translation().isEmpty() )
untranslated.insert( zeroKey((*it).sourceText()), *it );
} else if ( !(*it).translation().isEmpty() ) {
translated.insert( zeroKey((*it).sourceText()), *it );
}
}
for ( u = untranslated.begin(); u != untranslated.end(); ++u ) {
t = translated.find( u.key() );
if ( t != translated.end() && !t.key().isEmpty() &&
qstrcmp((*t).sourceText(), (*u).sourceText()) != 0 ) {
MetaTranslatorMessage m( *u );
m.setTranslation( translationAttempt((*t).translation(),
(*t).sourceText(),
(*u).sourceText()) );
tor->insert( m );
inserted++;
}
}
if ( verbose && inserted != 0 )
qWarning( " number heuristic provided %d translation%s",
inserted, inserted == 1 ? "" : "s" );
}